Assessment# 118115, Question# 129773, Content# 129765

$42\\ " *(1\\="" foot="" (12\\="" inches="" )="" 1="" 2\\="" feet="$

Assessment# 53398, Question# 53410, Content# 53404

You may be confusing Celsius and Fahrenheit. While $\displaystyle{{120}}^{\circ}\text{C}$ would be above the boiling point of water (which is $\displaystyle{{100}}^{\circ}\text{C}$ ), $\displaystyle{{120}}^{\circ}\text{F}$ is well below the boiling point of water (which is $\displaystyle{{212}}^{\circ}\text{F}$ ).

Assessment# 37106, Question# 37136, Content# 37141

The cost of the down payment is $1/6*" \\$="" 1500="\\$" 250="$ . The amount left over is $" 1500-="" \\$="" 250="\\$" 1250="$ .

Assessment# 37106, Question# 37136, Content# 37145

This is the correct fraction that is left over, not the amount in dollars. You could find the answer by multiplying this fraction by the total amount: $5/6*" \\$="" 1500="$ . The correct answer is $1250.

Assessment# 37106, Question# 37136, Content# 37147

You may have found $\displaystyle\frac{{1}}{{5}}$ of the total cost instead of $\displaystyle\frac{{1}}{{6}}$ . The cost of the down payment is $1/6*" \\$="" 1500="\\$" 250="$ . The correct answer is $1250.

Assessment# 37108, Question# 37199, Content# 37204

The amount of money the students have raised so far is $2200*3/5=6600/5=" \\$="" 1320="$ . They still need to raise $" 2200-="" \\$="" 1320="\\$" 880="$ .

Assessment# 37108, Question# 37199, Content# 37208

This is the fraction that they still need to raise, not the amount in dollars. You could find the answer by multiplying this fraction by the total amount: $" \\$="" 2200*2="" 5="$ . The correct answer is $880.

Assessment# 37108, Question# 37199, Content# 37210

You may have made an error in your calculations. The amount of money the students have raised so far is $" \\$="" 2200*3="" 5="$ . Subtract this product from the total to find the amount they still need to raise . The correct answer is $880.

Assessment# 29393, Question# 38458, Content# 38463

$\displaystyle\frac{\text{dividend}}{\text{divisor}}=\text{quotient}$

Assessment# 49457, Question# 49495, Content# 49506

You may have been thinking of $\displaystyle\frac{{3}}{{8}}$ quart. Multiply by $\displaystyle\frac{{{16}\ \text{ cups}}}{{{1}\ \text{ gallon}}}$ .

The correct answer is 6 cups.

Assessment# 45238, Question# 45240, Content# 45245

$\displaystyle{4}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={48}\ \text{ inches}$

Assessment# 45238, Question# 45240, Content# 45247

You may have multiplied 4 feet by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ . You should multiply 4 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ .

Assessment# 45112, Question# 45166, Content# 45171

$\displaystyle\frac{{{n}}}{{{36}\ \text{ attempts}}}=\frac{{{5}\ \text{ made}}}{{{6}\ \text{ attempts}}}$

$\displaystyle{n}\cdot{6}={36}\cdot{5}$

$\displaystyle{n}=\frac{{180}}{{6}}$

$\displaystyle{n}={30}$

Assessment# 45112, Question# 45142, Content# 45165

Be careful to use the correct units. The correct proportion is $\displaystyle\frac{{{2}\ \text{ horn players}}}{{{5}\ \text{ string players}}}=\frac{{{n}}}{{{40}\ \text{ string players}}}$ . The answer is 16 horn players.

Assessment# 45112, Question# 45142, Content# 45159

$\displaystyle\frac{{{2}\ \text{ horn players}}}{{{5}\ \text{ string players}}}=\frac{{{n}}}{{{40}\ \text{ string players}}}$

$\displaystyle{2}\cdot{40}={5}\cdot{n}$

$\displaystyle\frac{{80}}{{5}}={n}$

$\displaystyle{n}={16}$

Assessment# 45112, Question# 45126, Content# 45131

$\displaystyle\frac{{{60}\ \text{ words}}}{{{1}\ \text{ minute}}}=\frac{{{720}\ \text{ words}}}{{{n}}}$

$\displaystyle{60}\cdot{n}={720}\cdot{1}$

$\displaystyle{n}=\frac{{720}}{{60}}={12}$

Assessment# 45112, Question# 45114, Content# 45119

Set up a proportion. Cross multiply and use division to find the unknown value.

$\displaystyle\frac{{{1}\ \text{ student with pet}}}{{{2}\ \text{ students}}}=\frac{{{n}\ \text{ students with pets}}}{{{30}\ \text{ students}}}$

$\displaystyle{2}{n}={30}$

$\displaystyle{n}={15}$

Assessment# 53398, Question# 53410, Content# 53402

The boiling point of water is $\displaystyle{{100}}^{\circ}$ Celsius, and $\displaystyle{{95}}^{\circ}\text{C}$ is just below this temperature, so this could be the temperature of the water.

Assessment# 42528, Question# 42535, Content# 42540

$0.725 = 72.5" \\%="$ and $\displaystyle{0.725}=\frac{{725}}{{1000}}=\frac{{29}}{{40}}$

Assessment# 49457, Question# 49495, Content# 49500

$\displaystyle\frac{{3}}{{8}}\ \text{ gallon}=\frac{{{3}\ \text{ gallons}}}{{{8}}}\cdot\frac{{{16}\ \text{ cups}}}{{{1}\ \text{ gallon}}}={6}\ \text{ cups}$

Assessment# 49457, Question# 49495, Content# 49502

You may have converted to the number of ounces. Multiply by $\displaystyle\frac{{{16}\ \text{ cups}}}{{{1}\ \text{ gallon}}}$ .

The correct answer is 6 cups.

Assessment# 43042, Question# 43049, Content# 43054

The relationship can be written as the fraction $(12\\ " pages="" (80\\="" minutes="" )="$ . This can be simplified by dividing the numerator and denominator by 4.

Assessment# 43091, Question# 43098, Content# 43103

The rate as a fraction is $\displaystyle\frac{{{126}\ \text{ stairs}}}{{{3}\ \text{ minutes}}}$ . The unit rate is $\displaystyle\frac{{{42}\ \text{ stairs}}}{{{1}\ \text{ minute}}}$ .

Assessment# 43091, Question# 43098, Content# 43107

Aliyah's rate is $\displaystyle\frac{{{126}\ \text{ stairs}}}{{{3}\ \text{ minutes}}}$ . When you simplify this, be careful not to switch the units of measure.

Assessment# 43091, Question# 43137, Content# 43142

The rate is $\displaystyle\frac{{{276}\ \text{ books}}}{{{6}\ \text{ shelves}}}$ . Divide the numerator and the denominator by 6. This gives you the unit rate of 46 books per shelf.

Assessment# 43091, Question# 43137, Content# 43144

This is the correct rate, but not the unit rate. Divide the numerator and the denominator by 6. The answer is $\displaystyle\frac{{{46}\ \text{ books}}}{{{1}\ \text{ shelf}}}$ .

Assessment# 43091, Question# 43137, Content# 43146

The rate is $\displaystyle\frac{{{276}\ \text{ books}}}{{{6}\ \text{ shelves}}}$ . When you simplify this, be careful not to switch the units of measure. The answer is $\displaystyle\frac{{{46}\ \text{ books}}}{{{1}\ \text{ shelf}}}$ .

Assessment# 43091, Question# 43137, Content# 43148

The problem compares quantities with different units, so it can be described as a rate. 46:1 is a ratio. The answer is $\displaystyle\frac{{{46}\ \text{ books}}}{{{1}\ \text{ shelf}}}$ .

Assessment# 43091, Question# 43151, Content# 43156

The rate is $\displaystyle\frac{{{279}\ \text{ feet}}}{{{9}\ \text{ seconds}}}$ . Express the fraction with 1 in the denominator to find the number of feet traveled in one second.

Assessment# 43091, Question# 43151, Content# 43160

The rate is $\displaystyle\frac{{{279}\ \text{ feet}}}{{{9}\ \text{ seconds}}}$ . When you simplify this, be careful not to switch the units of measure.

Assessment# 43181, Question# 43183, Content# 43190

The rate is $(" \\$="" 12)="" (4\\="" pounds="" )="$ . Express the fraction with 1 in the denominator by dividing the numerator and denominator by 4.

Assessment# 43122, Question# 43133, Content# 43208

6.98 rounds up to 7, and then $\displaystyle{75}\div{7}={10}\text{R}{5}$ . This means he has enough for 10 CDs, and not enough for 11 CDs.

Assessment# 43181, Question# 44874, Content# 44879

The Spring Water rate is $(" 1.76)="" (16="" ounces="" ounce="" )="(" \\$="" 0.11)="" (1="$

and the Fresh Water rate is $(" 2.40)="" (20="" ounces="" ounce="" )="(" \\$="" 0.12)="" (1="$ .

Assessment# 45238, Question# 45240, Content# 45249

You may have multiplied 4 feet by $\displaystyle\frac{{{3}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ . You should multiply 4 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ .

Assessment# 45238, Question# 45240, Content# 45251

You may have multiplied 4 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ . You should multiply 4 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ .

Assessment# 45238, Question# 45276, Content# 45281

$\displaystyle{66}\ \text{ inches}\cdot\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}={5}\frac{{1}}{{2}}\ \text{ feet}$

Assessment# 45238, Question# 45276, Content# 45283

You may have been thinking of converting feet to yards and multiplied by $\displaystyle\frac{{1}}{{3}}$ . You need to multiply by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 45238, Question# 45276, Content# 45285

You may have multiplied 66 inches by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ . You need to multiply 66 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 45238, Question# 45276, Content# 45287

You may have multiplied 66 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{10}\ \text{ inches}}}$ . You need to multiply 66 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 52711, Question# 53575, Content# 53580

There are 2,000 pounds in a ton.

$\displaystyle{2},{000}\ \text{ pounds}-{500}\ \text{ pounds}={1},{500}\ \text{ pounds}$

Assessment# 42528, Question# 46226, Content# 46231

$7/20=35/100=0.35=35" \\%="$

Assessment# 46238, Question# 46240, Content# 46245

$\displaystyle{1}\ \text{ foot}={12}\ \text{ inches}$ and $\displaystyle{12}\ \text{ inches}+{9}\ \text{ inches}={21}\ \text{ inches}$ .

Assessment# 46238, Question# 46240, Content# 46247

You may have thought that $\displaystyle{1}\ \text{ foot}={10}\ \text{ inches}$ . You need to use $\displaystyle{1}\ \text{ foot}={12}\ \text{ inches}$ and then add 12 inches and 9 inches.

Assessment# 46238, Question# 46240, Content# 46249

You may have thought that $\displaystyle{1}\ \text{ foot}={3}\ \text{ inches}$ . You need to use $\displaystyle{1}\ \text{ foot}={12}\ \text{ inches}$ and then add 12 inches and 9 inches.

Assessment# 46238, Question# 46271, Content# 46276

$\displaystyle{4}\ \text{ yards}\cdot\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}={12}\ \text{ feet}$ and $\displaystyle{12}\ \text{ feet}+{2}\ \text{ feet}={14}\ \text{ feet}$ .

Assessment# 46238, Question# 46271, Content# 46278

You may have multiplied 4 yards by $\displaystyle\frac{{{10}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then added 2 feet. You need to multiply 4 yards by $\displaystyle\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then add 2 feet. The correct answer is 14 feet.

Assessment# 46238, Question# 46271, Content# 46280

You may have multiplied 4 yards by $\displaystyle\frac{{{12}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then added 2 feet. You need to multiply 4 yards by $\displaystyle\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then add 2 feet. The correct answer is 14 feet.

Assessment# 46238, Question# 46271, Content# 46282

You may have multiplied 4 yards by $\displaystyle\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ , but then forgot to add 2 feet. The correct answer is 14 feet.

Assessment# 46238, Question# 46283, Content# 46288

$\displaystyle{3}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={36}\ \text{ inches}$ and $\displaystyle{36}\ \text{ inches}+{5}\ \text{ inches}={41}\ \text{ inches}$ .

Assessment# 46238, Question# 46283, Content# 46290

You may have multiplied 3 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 5 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 5 inches.

Assessment# 46238, Question# 46283, Content# 46292

You may have multiplied 3 feet by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ and then added 5 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 5 inches.

Assessment# 46295, Question# 46297, Content# 46302

$\displaystyle{4}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={48}\ \text{ inches}$ and $\displaystyle{48}+{10}={58}$ inches.

Assessment# 46295, Question# 46297, Content# 46306

You may have multiplied 4 feet by $\displaystyle\frac{{{3}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 10 inches. You need to multiply 4 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 10 inches.

Assessment# 46295, Question# 46297, Content# 46308

You may have multiplied 4 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 10 inches. You need to multiply 4 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 10 inches.

Assessment# 46295, Question# 46314, Content# 46319

$(" 16.50)="" (10="" yards="" -:="" 10)="" yard="" )="(" \\$="" 16.50="$ and $(" \\$="" 1.65)="" (1\\="" )*(1\\="" yard="" (3\\="" feet="" foot="" )="$

Assessment# 46295, Question# 46314, Content# 46321

You may have forgotten to convert the cost per yard to the cost per foot. You need to multiply the cost per yard by $\displaystyle\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}$ .

Assessment# 46295, Question# 46314, Content# 46323

You may have multiplied the cost per yard by 3. You need to multiply the cost per yard by $\displaystyle\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}$ .

Assessment# 46295, Question# 46314, Content# 46325

You may have just divided $16.50 by 3. You need to divide $16.50 by 10 to find the cost per yard and then multiply the result by $\displaystyle\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}$ .

Assessment# 46295, Question# 46326, Content# 46331

$\displaystyle{18}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={216}\ \text{ inches}$ ; $\displaystyle{12}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={144}\ \text{ inches}$ ; $\displaystyle{144}\ \text{ inches}+{9}\ \text{ inches}={153}\ \text{ inches}$ ; $\displaystyle{216}\ \text{ inches}-{153}\ \text{ inches}={63}\ \text{ inches}$ .

Assessment# 46295, Question# 46338, Content# 46343

$\displaystyle{200}\ \text{ yards}\cdot\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}={600}\ \text{ feet}$ . To find the perimeter of a rectangle, you can use the formula $\displaystyle\text{perimeter}=\text{length}+\text{length}+\text{width}+\text{width}$ . $\displaystyle{600}\ \text{ feet}+{600}\ \text{ feet}+{300}\ \text{ feet}+{300}\ \text{ feet}={1},{800}\ \text{ feet}$ .

Assessment# 48394, Question# 48401, Content# 48406

Because $\displaystyle\text{percent}\cdot\text{base}=\text{amount}$ , the equation for this problem is $\displaystyle{n}\cdot{56}={14}$ . The corresponding division is $\displaystyle{14}\div{56}$ , so $\displaystyle{n}={0.25}$ . Rewriting this as a percent gives 25%.

Assessment# 48726, Question# 48767, Content# 48759

The relationship can be written as the fraction $\displaystyle\frac{{{4}\ \text{ miles}}}{{{74}\ \text{ minutes}}}$ . This can be simplified by dividing the numerator and denominator by 2.

Assessment# 48781, Question# 48810, Content# 48802

The rate is $\displaystyle\frac{{{288}\ \text{ seats}}}{{{9}\ \text{ rows}}}$ . Divide the numerator and the denominator by 9. This gives you the unit rate of 32 seats per row.

Assessment# 48781, Question# 48810, Content# 48804

This is the correct rate, but not the unit rate. Divide the numerator and the denominator by 9. The correct answer: $\displaystyle\frac{{{32}\ \text{ seats}}}{{{1}\ \text{ row}}}$

Assessment# 48781, Question# 48810, Content# 48806

The rate is $\displaystyle\frac{{{288}\ \text{ seats}}}{{{9}\ \text{ rows}}}$ . When you simplify this rate, be careful not to switch the units of measure. The correct answer: $\displaystyle\frac{{{32}\ \text{ seats}}}{{{1}\ \text{ row}}}$

Assessment# 48781, Question# 48810, Content# 48808

The problem compares quantities with different units, so it can be described as a rate. 32:1 is a ratio. The correct answer: $\displaystyle\frac{{{32}\ \text{ seats}}}{{{1}\ \text{ row}}}$

Assessment# 48781, Question# 48822, Content# 48814

The rate is $\displaystyle\frac{{{375}\ \text{ bottles}}}{{{5}\ \text{ minutes}}}$ . Express the fraction with 1 in the denominator to find the number of bottles per minute.

Assessment# 48781, Question# 48822, Content# 48818

The rate is $\displaystyle\frac{{{375}\ \text{ bottles}}}{{{5}\ \text{ minutes}}}$ . When you simplify this rate, be careful not to switch the units of measure.

Assessment# 48781, Question# 48834, Content# 48826

The rate as a fraction is $\displaystyle\frac{{{124}\ \text{ pages}}}{{{62}\ \text{ minutes}}}$ . The unit rate is $\displaystyle\frac{{{2}\ \text{ pages}}}{{{1}\ \text{ minute}}}$ .

Assessment# 48781, Question# 48834, Content# 48830

Jim's rate is $\displaystyle\frac{{{124}\ \text{ pages}}}{{{62}\ \text{ minutes}}}$ . When you simplify this, be careful not to switch the units of measure.

Assessment# 48847, Question# 48865, Content# 48857

The Kitty Kibbles rate is $(" 20.80)="" (16="" pounds="" pound="" )="(" \\$="" 1.30)="" (1="$

and the Feline Flavor rate is $(" 11.20)="" (8="" pounds="" pound="" )="(" \\$="" 1.40)="" (1="$ .

Assessment# 48847, Question# 48877, Content# 48869

The rate is $(" \\$="" 25)="" (5="" pounds="" )="$ . Express the fraction with 1 in the denominator by dividing the numerator and denominator by 5.

Assessment# 48900, Question# 48926, Content# 48918

$42\\ " *(1\\="" foot="" (12\\="" inches="" )="" 1="" 2\\="" feet="$

Assessment# 48900, Question# 48926, Content# 48920

Assessment# 48900, Question# 48926, Content# 48922

You may have multiplied 42 inches by $(12\\ " inches="" (1\\="" foot="" )="$ . You need to multiply 42 inches by $(1\\ " foot="" (12\\="" inches="" )="$ to get the correct answer.

Assessment# 48900, Question# 48926, Content# 48924

You may have multiplied 42 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{10}\ \text{ inches}}}$ . You need to multiply 42 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 48900, Question# 48949, Content# 48941

$5\\ " feet="" *(12\\="" (1\\="" foot="" )="" inches="$

Assessment# 48900, Question# 48949, Content# 48943

You may have multiplied 5 feet by $(1\\ " foot="" (12\\="" inches="" )="$ . You should multiply 5 feet by $(12\\ " inches="" (1\\="" foot="" )="$ .

Assessment# 48900, Question# 48949, Content# 48945

You may have multiplied 5 feet by $(3\\ " inches="" (1\\="" foot="" )="$ . You should multiply 5 feet by $(12\\ " inches="" (1\\="" foot="" )="$ .

Assessment# 48900, Question# 48949, Content# 48947

You may have multiplied 5 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ . You should multiply 5 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ .

Assessment# 48963, Question# 48975, Content# 48967

$4\\ " feet="" *(12\\="" (1\\="" foot="" )="" inches="$ and $48\\ " +3\\="" inches="$

Assessment# 48963, Question# 48975, Content# 48969

You may have multiplied 4 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 3 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 3 inches.

Assessment# 48963, Question# 48975, Content# 48971

You may have multiplied 4 feet by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ and then added 3 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 3 inches.

Assessment# 48963, Question# 48993, Content# 48985

$5\\ " yards="" *(3\\="" (1\\="" yard="" )="" feet="$ and $\displaystyle{15}\ \text{ feet}+{1}\ \text{ foot}={16}\ \text{ feet}$

Assessment# 48963, Question# 48993, Content# 48987

You may have multiplied 5 yards by $\displaystyle\frac{{{10}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then added 1 foot. You need to multiply 5 yards by $\displaystyle\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then add 1 foot. The correct answer is 16 feet.

Assessment# 48963, Question# 48993, Content# 48989

You may have multiplied 5 yards by $\displaystyle\frac{{{12}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then added 1 foot. You need to multiply 5 yards by $\displaystyle\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ and then add 1 foot. The correct answer is 16 feet.

Assessment# 48963, Question# 48993, Content# 48991

You may have correctly multiplied 5 yards by $\displaystyle\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}$ , but then you may have forgotten to add 1 foot. The correct answer is 16 feet.

Assessment# 48963, Question# 49022, Content# 49016

$\displaystyle{1}\ \text{ foot}={12}\ \text{ inches}$ and $\displaystyle{12}\ \text{ inches}+{3}\ \text{ inches}={15}\ \text{ inches}$ .

Assessment# 48963, Question# 49022, Content# 49018

You may have thought that $\displaystyle{1}\ \text{ foot}={10}\ \text{ inches}$ . You need to use $\displaystyle{1}\ \text{ foot}={12}\ \text{ inches}$ and then add 12 inches and 3 inches.

Assessment# 48963, Question# 49022, Content# 49020

You may have thought that $\displaystyle{1}\ \text{ foot}={3}\ \text{ inches}$ . You need to use $\displaystyle{1}\ \text{ foot}={12}\ \text{ inches}$ and then add 12 inches and 3 inches.

Assessment# 49095, Question# 49115, Content# 49107

$31\\ " feet="" *(12\\="" (1\\="" foot="" )="" inches="$

$\displaystyle{17}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={204}\ \text{ inches}$

$204\\ " +="" 4\\="" inches="$

$372\\ " -="" 208\\="" inches="$

Assessment# 49095, Question# 49127, Content# 49119

$180\\ " yards="" *(3\\="" (1\\="" yard="" )="" feet="$

To find the perimeter of a rectangle, you can use the formula $\displaystyle\text{perimeter}=\text{length}+\text{length}+\text{width}+\text{width}$ .

$\displaystyle{540}\ \text{ feet}+{540}\ \text{ feet}+{200}\ \text{ feet}+{200}\ \text{ feet}={1480}\ \text{ feet}$

Assessment# 49095, Question# 49127, Content# 49121

You may have converted the 200 feet to $\displaystyle{66}\frac{{2}}{{3}}\ \text{ yards}$ , then found the total perimeter in yards. You need to convert the 180 yards to feet, then find the total perimeter.

Assessment# 49095, Question# 49139, Content# 49131

$(" 15.30)="" yards="" -:10)="(" 1.53)="" (1\\="" yard="" )="(" \\$="" 15.30-:10)="" (10\\="$ and $(" \\$="" \\1.53)="" (1\\="" )*(1\\="" yard="" (3\\="" feet="" )="$

Assessment# 49095, Question# 49139, Content# 49133

You may have forgotten to convert the cost per yard to the cost per foot. You need to multiply the cost per yard by $\displaystyle\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}$ .

Assessment# 49095, Question# 49139, Content# 49135

You may have multiplied the cost per yard by 3. You need to multiply the cost per yard by $\displaystyle\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}$ to get the correct answer.

Assessment# 49095, Question# 49139, Content# 49137

You may have just divided $15.30 by 3. You need to divide $15.30 by 10 to find the cost per yard and then multiply the result by $\displaystyle\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}$ to get the correct answer.

Assessment# 49095, Question# 49151, Content# 49143

$3\\ " feet="" *(12\\="" (1\\="" foot="" )="" inches="$ and $\displaystyle{36}+{6}={42}$ inches

Assessment# 49095, Question# 49151, Content# 49147

You may have multiplied 3 feet by $\displaystyle\frac{{{3}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 6 inches. You need to multiply 3 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 6 inches.

Assessment# 49095, Question# 49151, Content# 49149

You may have multiplied 3 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 6 inches. You need to multiply 3 feet by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 6 inches.

Assessment# 49457, Question# 49507, Content# 49512

$\displaystyle{2}\ \text{ pints}\cdot\frac{{{2}\ \text{ cups}}}{{{1}\ \text{ pint}}}\cdot\frac{{{8}\ \text{ ounces}}}{{{1}\ \text{ cup}}}={32}\ \text{ ounces}$

Assessment# 49519, Question# 49531, Content# 49523

$\displaystyle\frac{{{3}\ \text{ SUVs}}}{{{10}\ \text{ cars}}}=\frac{{{n}\ \text{ SUVs}}}{{{330}\ \text{ cars}}}$

$\displaystyle{3}\cdot{330}={10}\cdot{n}$

$\displaystyle\frac{{990}}{{10}}={n}$

$\displaystyle{n}={99}$

Assessment# 49519, Question# 49531, Content# 49525

You may be thinking of the ratio of 330 cars to 10 cars, but this does not tell you how many SUVs there are. You need to set up a proportion and solve for the unknown number of SUVs. The correct proportion is $\displaystyle\frac{{{3}\ \text{ SUVs}}}{{{10}\ \text{ cars}}}=\frac{{{n}\ \text{ SUVs}}}{{{330}\ \text{ cars}}}$ . The answer is 99 SUVs.

Assessment# 49519, Question# 49531, Content# 49529

You may have divided 330 by 30. You need to set up a proportion and solve for the unknown number of SUVs. The correct proportion of SUVs to cars is $\displaystyle\frac{{{3}\ \text{ SUVs}}}{{{10}\ \text{ cars}}}=\frac{{{n}\ \text{ SUVs}}}{{{330}\ \text{ cars}}}$ . The answer is 99 SUVs.

Assessment# 49519, Question# 49543, Content# 49535

$\displaystyle\frac{{{70}\ \text{ words}}}{{{1}\ \text{ minute}}}=\frac{{{n}\ \text{ words}}}{{{15}\ \text{ minutes}}}$

$\displaystyle{1}\cdot{n}={70}\cdot{15}$

$\displaystyle{n}={1},{050}$

Assessment# 49519, Question# 49555, Content# 49547

Set up a proportion. Cross multiply and use division to find the unknown value.

$\displaystyle\frac{{{1}\ \text{ graphic novel}}}{{{5}\ \text{ books}}}=\frac{{{n}\ \text{ graphic novels}}}{{{50}\ \text{ books}}}$

$\displaystyle{5}{n}={50}$

$\displaystyle{n}={10}$

Assessment# 49947, Question# 49961, Content# 49966

Convert both amounts to gallons and add:

$\displaystyle\frac{{{2}\ \text{ quarts}}}{{{1}}}\cdot\frac{{{1}\ \text{ gallon}}}{{{4}\ \text{ quarts}}}=\frac{{1}}{{2}}\ \text{ gallon}$

$\displaystyle\frac{{{8}\ \text{ pints}}}{{{1}}}\cdot\frac{{{1}\ \text{ quart}}}{{{2}\ \text{ pints}}}\cdot\frac{{{1}\ \text{ gallon}}}{{{4}\ \text{ quarts}}}={1}\ \text{ gallon}$

$\displaystyle\frac{{1}}{{2}}+{1}={1}\frac{{1}}{{2}}\ \text{ gallons}$

Assessment# 49519, Question# 49567, Content# 49559

$\displaystyle\frac{{{4}\ \text{ blocks}}}{{{5}\ \text{ kicks}}}=\frac{{{n}\ \text{ blocks}}}{{{20}\ \text{ kicks}}}$

$\displaystyle{4}\cdot{20}={5}\cdot{n}$

$\displaystyle\frac{{80}}{{5}}={n}$

$\displaystyle{n}={16}$

Assessment# 49947, Question# 49949, Content# 49954

$\displaystyle\frac{{{40}\ \text{ cups}}}{{{1}}}\cdot\frac{{{1}\ \text{ gallon}}}{{{16}\ \text{ cups}}}={2.5}\ \text{ gallons}$

The amount left over is $\displaystyle{6}-{2.5}={3.5}$ gallons.

Assessment# 52711, Question# 52713, Content# 52718

$\displaystyle{8},{000}\ \text{ pounds}\cdot\frac{{{1}\ \text{ ton}}}{{{2},{000}\ \text{ pounds}}}={4}\ \text{ tons}$ and $4 * " \\$="" 90="\\$" 360="$

Assessment# 49947, Question# 49961, Content# 49968

You may have thought that there were 2 quarts in one gallon. There are 4 quarts in one gallon. The correct answer is $\displaystyle{1}\frac{{1}}{{2}}\ \text{ gallons}$ .

Assessment# 49947, Question# 49961, Content# 49970

You may have confused pints and cups, and incorrectly converted 8 pints to $\displaystyle\frac{{1}}{{2}}\ \text{ gallon}$ . The correct answer is $\displaystyle{1}\frac{{1}}{{2}}\ \text{ gallons}$ .

Assessment# 49947, Question# 49961, Content# 49972

You may have found the total for 8 quarts and 2 pints, instead of the given 2 quarts and 8 pints. The correct answer is $\displaystyle{1}\frac{{1}}{{2}}\ \text{ gallons}$ .

Assessment# 49947, Question# 49973, Content# 49978

He needs 20 pints.

$\displaystyle\frac{{{3}\ \text{ gallons}}}{{{1}}}\cdot\frac{{{4}\ \text{ quarts}}}{{{1}\ \text{ gallon}}}\cdot\frac{{{2}\ \text{ pints}}}{{{1}\ \text{ quart}}}={24}\ \text{ pints}$ , while

$\displaystyle\frac{{{2}\ \text{ gallons}}}{{{1}}}\cdot\frac{{{4}\ \text{ quarts}}}{{{1}\ \text{ gallon}}}\cdot\frac{{{2}\ \text{ pints}}}{{{1}\ \text{ quart}}}={16}\ \text{ pints}$ .

Assessment# 50005, Question# 50041, Content# 50033

$13/25=(13*4)/(25*4)=52/100=0.52=52" \\%="$

Assessment# 50005, Question# 50053, Content# 50045

$28" \\%="$ and $\displaystyle{0.28}=\frac{{28}}{{100}}=\frac{{7}}{{25}}$

Assessment# 50115, Question# 50163, Content# 50155

Because $\displaystyle\text{percent}\cdot\text{base}=\text{amount}$ , the equation for this problem is $\displaystyle{0.65}\cdot{n}={52}$ . The corresponding division is $\displaystyle{52}\div{0.65}$ , so $\displaystyle{n}={80}$ .

Assessment# 50115, Question# 50172, Content# 50177

The amount is 8, the base is 20, and the percent is unknown. Use the equation $\displaystyle\text{percent}=\frac{{\text{amount}}}{{\text{base}}}$ . The left side is $\displaystyle\frac{{n}}{{100}}$ , and the right side is $\displaystyle\frac{{8}}{{20}}$ .

Assessment# 50115, Question# 50172, Content# 50179

If you use the equation $\displaystyle\text{percent}=\frac{{\text{amount}}}{{\text{base}}}$ , the left side is correct, but you have confused the amount and the base.

Assessment# 50115, Question# 50172, Content# 50183

If you use the equation $\displaystyle\text{percent}=\frac{{\text{amount}}}{{\text{base}}}$ , then you have written a proportion with 8% being equal to an unknown amount over 20. The unknown in this problem is the percent.

Assessment# 52457, Question# 52469, Content# 52474

$\displaystyle\frac{{{0.6}\ \text{ liters}}}{{{1}}}\cdot\frac{{{1000}\ \text{ milliliters}}}{{{1}\ \text{ liter}}}={600}\ \text{ milliliters}$ ; $\displaystyle\frac{{{1.25}\ \text{ deciliters}}}{{{1}}}\cdot\frac{{{100}\ \text{ milliliters}}}{{{1}\ \text{ deciliter}}}={125}\ \text{ milliliters}$ ;

$\displaystyle{600}-{125}={475}$ milliliters

Assessment# 52517, Question# 52543, Content# 52548

$\displaystyle{448},{000}\ \text{ ounces}\cdot\frac{{{1}\ \text{ pound}}}{{{16}\ \text{ ounces}}}\cdot\frac{{{1}\ \text{ ton}}}{{{2},{000}\ \text{ pounds}}}={14}\ \text{ tons}$

Assessment# 52517, Question# 52531, Content# 52536

$\displaystyle{22},{000}\ \text{ pounds}\cdot\frac{{{1}\ \text{ ton}}}{{{2},{000}\ \text{ pounds}}}={11}\ \text{ tons}$

Assessment# 52517, Question# 52519, Content# 52524

$\displaystyle{120}\ \text{ ounces}\cdot\frac{{{1}\ \text{ pound}}}{{{16}\ \text{ ounces}}}={7.5}\ \text{ pounds}$

Assessment# 52457, Question# 52481, Content# 52486

$\displaystyle{1.8}\ \text{ kilograms}={1},{800}\ \text{ grams}$ and $\displaystyle{315}\ \text{ dekagrams}={3},{150}\ \text{ grams}$ , so the total weight of the apples and oranges was $\displaystyle{1},{800}+{3},{150}={4},{950}$ grams. The weight of the bananas was $\displaystyle{4},{950}-{1},{350}={3},{600}$ grams.

Assessment# 52457, Question# 52481, Content# 52492

You may have correctly done the conversion $\displaystyle{1.8}\ \text{ kilograms}={1},{800}\ \text{ grams}$ , but forgot to convert the 315 dekagrams to grams when adding the weights of the apples and the oranges.

Assessment# 52457, Question# 52493, Content# 52498

The distance between the cities is $\displaystyle\frac{{{280},{000}\ \text{ meters}}}{{{1}}}\cdot\frac{{{1}\ \text{ kilometer}}}{{{1},{000}\ \text{ meters}}}$ , or 280 kilometers.

$\displaystyle\frac{{{280}\ \text{ kilometers}}}{{{n}}}=\frac{{{80}\ \text{ kilometers}}}{{{1}\ \text{ hour}}}$

$\displaystyle{n}=\frac{{280}}{{80}}={3}\frac{{1}}{{2}}$

Assessment# 52457, Question# 52493, Content# 52504

You may not have converted correctly from meters to kilometers. $\displaystyle\frac{{{280},{000}\ \text{ meters}}}{{{1}}}\cdot\frac{{{1}\ \text{ kilometer}}}{{{1},{000}\ \text{ meters}}}={280}\ \text{ kilometers}$ . The correct answer is $\displaystyle{3}\frac{{1}}{{2}}$ hours.

Assessment# 52457, Question# 52493, Content# 52502

You may have correctly converted the distance between cities to 280 kilometers, and then solved the proportion $\displaystyle\frac{{80}}{{n}}=\frac{{280}}{{1}}$ . The correct equation would be $\displaystyle\frac{{{280}\ \text{ kilometers}}}{{{n}}}=\frac{{{80}\ \text{ kilometers}}}{{{1}\ \text{ hour}}}$ . The correct answer is $\displaystyle{3}\frac{{1}}{{2}}$ hours.

Assessment# 52457, Question# 52505, Content# 52510

$\displaystyle\frac{{{7.47}\ \text{ hectograms}}}{{{1}}}\cdot\frac{{{1}\ \text{ kilogram}}}{{{10}\ \text{ hectograms}}}={0.747}\ \text{ kilogram}$

$\displaystyle{0.766}\ \text{ kg}-{0.747}\ \text{ kg}={0.029}\ \text{ kg}$

Assessment# 52711, Question# 52725, Content# 52730

The total weight is 8 pounds. The first 4 cost $1.50 and the other 4 cost $" \\$="" 1xx4="\\$" 4="$ . Together: $5.50

Assessment# 52711, Question# 52737, Content# 52742

$\displaystyle{3},{200}\ \text{ pounds}+{5},{100}\ \text{ pounds}+{700}\ \text{ pounds}={9},{000}\ \text{ pounds}$ . Convert pounds to tons:

$\displaystyle{9},{000}\ \text{ pounds}\cdot\frac{{{1}\ \text{ ton}}}{{{2},{000}\ \text{ pounds}}}={4}\frac{{1}}{{2}}\ \text{ tons}$

Assessment# 49947, Question# 52768, Content# 52773

He needs 48 cups of tea. Converting cups to gallons:

$\displaystyle\frac{{{48}\ \text{ cups}}}{{{1}}}\cdot\frac{{{1}\ \text{ gallon}}}{{{16}\ \text{ cups}}}={3}\ \text{ gallons}$

Assessment# 52776, Question# 52788, Content# 52780

$6\\ " quarts="" *(4\\="" cups="" quart="" )*(8\\="" (1\\="" cup="" )="" fluid="" ounces="$

Assessment# 53398, Question# 53410, Content# 53408

You may be confusing Celsius and Fahrenheit. While $\displaystyle{{99}}^{\circ}\text{C}$ would be close to the boiling point of water (which is $\displaystyle{{100}}^{\circ}\text{C}$ ), $\displaystyle{{99}}^{\circ}\text{F}$ is far below the boiling point of water (which is $\displaystyle{{212}}^{\circ}\text{F}$ ).

Assessment# 53398, Question# 53422, Content# 53416

This is the correct freezing point of water in degrees Celsius, but it is the boiling point of water in degrees Fahrenheit. The correct answer is $\displaystyle{{0}}^{\circ}\text{C}$ and $\displaystyle{{32}}^{\circ}\text{F}$ .

Assessment# 53398, Question# 53422, Content# 53418

This is the correct freezing point of water in degrees Celsius, but 100 degrees would be the boiling point in degrees Celsius. Thec orrect answer is $\displaystyle{{0}}^{\circ}\text{C}$ and $\displaystyle{{32}}^{\circ}\text{F}$ .

Assessment# 53036, Question# 53501, Content# 53506

This temperature is below the freezing point of water, which is $\displaystyle{{32}}^{\circ}\text{F}$ . Therefore, the lake could be frozen.

Assessment# 53036, Question# 53501, Content# 53508

You may be confusing Celsius and Fahrenheit. While $\displaystyle{{20}}^{\circ}\text{F}$ would be below the freezing point of water (which is $\displaystyle{{32}}^{\circ}\text{F}$ ), $\displaystyle{{20}}^{\circ}\text{C}$ is above the freezing point of water (which is $\displaystyle{{0}}^{\circ}\text{C}$ ). The correct answer is $\displaystyle{{30}}^{\circ}\text{F}$ .

Assessment# 53036, Question# 53501, Content# 53510

The freezing point of water is $\displaystyle{{32}}^{\circ}$ . Because $\displaystyle{{35}}^{\circ}$ is above this, the water could not be frozen. The correct answer is $\displaystyle{{30}}^{\circ}\text{F}$ .

Assessment# 53036, Question# 53501, Content# 53512

You may be confusing Celsius and Fahrenheit. While $\displaystyle{{25}}^{\circ}\text{F}$ would be below the freezing point of water (which is $\displaystyle{{32}}^{\circ}\text{F}$ ), $\displaystyle{{25}}^{\circ}\text{C}$ is above the freezing point of water (which is $\displaystyle{{0}}^{\circ}\text{C}$ ). The correct answer is $\displaystyle{{30}}^{\circ}\text{F}$ .

Assessment# 53398, Question# 53422, Content# 53420

These are the boiling points of water, not the freezing points. The correct answer is $\displaystyle{{0}}^{\circ}\text{C}$ and $\displaystyle{{32}}^{\circ}\text{F}$ .

Assessment# 52776, Question# 52814, Content# 52806

$4\\(1)/(2)\\ " quarts="" *(2\\="" (1\\="" quart="" )="" pints="$

Assessment# 52776, Question# 52814, Content# 52812

You may have been thinking of $\displaystyle{4}\frac{{{1}}}{{{2}}}$ gallons. Multiply: $4(1)/(2)\\ " quarts="" *(2\\="" pints="" (1\\="" quart="" )="$ . The correct answer is 9 pints.

Assessment# 52776, Question# 52826, Content# 52818

$\displaystyle{1}\ \text{ gallon}={4}\ \text{ quarts}={4}\ \text{ quarts}\cdot\frac{{{4}\ \text{ cups}}}{{{1}\ \text{ quart}}}={16}\ \text{ cups}$

Assessment# 49947, Question# 52839, Content# 52844

The bottles will be filled by $\displaystyle{96}\cdot{16}={1},{536}$ fluid ounces. Converting ounces to gallons:

$\displaystyle\frac{{{1536}\ \text{ fl oz}}}{{{1}}}\cdot\frac{{{1}\ \text{ cup}}}{{{8}\ \text{ fl oz}}}\cdot\frac{{{1}\ \text{ gallon}}}{{{16}\ \text{ cups}}}={12}\ \text{ gallons}$

Assessment# 52457, Question# 52505, Content# 52852

$\displaystyle\frac{{{7.47}\ \text{ hectograms}}}{{{1}}}\cdot\frac{{{1}\ \text{ kilogram}}}{{{10}\ \text{ hectograms}}}={0.747}\ \text{ kilogram}$

$\displaystyle{0.776}\ \text{ kg}-{0.747}\ \text{ kg}={0.029}\ \text{ kg}$

Assessment# 53047, Question# 53059, Content# 53051

$\displaystyle{0.25}\ \text{ dekaliters}={2.5}\ \text{ liters}$ and $\displaystyle{375}\ \text{ centiliters}={3.75}\ \text{ liters}$ , so the total volume of juices is $\displaystyle{2.5}+{3.75}={6.25}$ liters. $\displaystyle{6.25}-{1.125}={5.125}$ liters more juice than sparkling water.

Assessment# 53047, Question# 53059, Content# 53057

You may have correctly done the conversion $\displaystyle{375}\ \text{ centiliters}={3.75}\ \text{ liters}$ , but forgot to convert the 0.25 dekaliters to liters when adding the volumes of the juices.

Assessment# 53047, Question# 53071, Content# 53063

$\displaystyle\frac{{{7.19}\ \text{ meters}}}{{{1}}}\cdot\frac{{{100}\ \text{ centimeters}}}{{{1}\ \text{ meter}}}={719}\ \text{ centimeters}$ $\displaystyle\frac{{{6.9}\ \text{ meters}}}{{{1}}}\cdot\frac{{{100}\ \text{ centimeters}}}{{{1}\ \text{ meter}}}={690}\ \text{ centimeters}$

$\displaystyle{719}-{690}={29}$ centimeters

Assessment# 53047, Question# 53083, Content# 53075

$\displaystyle\frac{{{4},{930}\ \text{ centiliters}}}{{{1}}}\cdot\frac{{{1}\ \text{ liter}}}{{{100}\ \text{ centiliters}}}={49.3}\ \text{ liters}$

$\displaystyle{70.1}-{49.3}={20.8}$

Assessment# 53047, Question# 53095, Content# 53087

The distance between the cities is $\displaystyle\frac{{{921},{000}\ \text{ meters}}}{{{1}}}\cdot\frac{{{1}\ \text{ kilometer}}}{{{1},{000}\ \text{ meters}}}$ , or 921 kilometers.

$\displaystyle\frac{{{921}\ \text{ km}}}{{{n}}}=\frac{{{307}\ \text{ km}}}{{{1}\ \text{ hour}}}$

$\displaystyle{n}=\frac{{921}}{{307}}={3}$

Assessment# 53047, Question# 53095, Content# 53091

You may have correctly converted the distance between cities to 921 kilometers, and then solved the proportion $\displaystyle\frac{{307}}{{n}}=\frac{{921}}{{1}}$ . The correct equation would be $\displaystyle\frac{{{921}\ \text{ km}}}{{n}}=\frac{{{307}\ \text{ km}}}{{{1}\ \text{ hr}}}$ .

Assessment# 53047, Question# 53095, Content# 53093

You may not have converted correctly from meters to kilometers.

$\displaystyle\frac{{{921},{000}\ \text{ meters}}}{{{1}}}\cdot\frac{{{1}\ \text{ kilometer}}}{{{1},{000}\ \text{ meters}}}={921}\ \text{ km}$

Assessment# 53036, Question# 53131, Content# 53156

This is the correct freezing point of water in degrees Celsius, but it is the boiling point of water in degrees Fahreheit. The answer is $\displaystyle{{0}}^{\circ}\text{C}$ and $\displaystyle{{32}}^{\circ}\text{F}$ .

Assessment# 53036, Question# 53131, Content# 53158

This is the correct freezing point of water in degrees Celsius, but 100 degrees would be the boiling point in degrees Celsius. The answer is $\displaystyle{{0}}^{\circ}\text{C}$ and $\displaystyle{{32}}^{\circ}\text{F}$ .

Assessment# 53036, Question# 53131, Content# 53160

These are the boiling points of water, not the freezing points. The answer is $\displaystyle{{0}}^{\circ}\text{C}$ and $\displaystyle{{32}}^{\circ}\text{F}$ .

Assessment# 53173, Question# 53180, Content# 53185

Converting Montreal's temperature to Fahrenheit:

$\displaystyle{F}=\frac{{9}}{{5}}\cdot{20}+{32}$

$\displaystyle{F}=\frac{{180}}{{5}}+{32}={36}+{32}={68}$

$\displaystyle{{68}}^{\circ}\text{F}-{{65}}^{\circ}\text{F}={{3}}^{\circ}\text{F}$

Assessment# 53173, Question# 53180, Content# 53187

You may have mixed up which city had which temperature while calculating. New York's temperature is $\displaystyle{{65}}^{\circ}\text{F}$ . You need to convert Montreal's temperature to Fahrenheit to compare it with New York's temperature.

Assessment# 53212, Question# 53236, Content# 53228

$\displaystyle{6}\ \text{ pints}\cdot\frac{{{1}\ \text{ quart}}}{{{2}\ \text{ pints}}}={3}\ \text{ quarts}$ , and $\displaystyle{4}\ \text{ cups}={1}\ \text{ quart}$ , so the total is $\displaystyle{3}+{1}={4}$ quarts.

Assessment# 53212, Question# 53260, Content# 53252

$1\\ " quart="" pints="" cups="$ . And $4\\ " cups="" *(8\\="" (1\\="" cup="" )="" fluid="" ounces="$

Assessment# 53173, Question# 53206, Content# 53268

You used the wrong formula. To convert from Fahrenheit to Celsius, use the formula $\displaystyle{C}=\frac{{5}}{{9}}{\left({F}-{32}\right)}$ . The correct answer is $\displaystyle{{16}}^{\circ}\text{C}$ .

Assessment# 53173, Question# 53206, Content# 53270

You used the ratio $\displaystyle\frac{{9}}{{5}}$ in the conversion formula, instead of $\displaystyle\frac{{5}}{{9}}$ . The correct answer is $\displaystyle{{16}}^{\circ}\text{C}$ .

Assessment# 53173, Question# 53206, Content# 53272

You added 32 to the degrees Fahrenheit in the conversion formula, instead of subtracting 32. The correct answer is $\displaystyle{{16}}^{\circ}\text{C}$ .

Assessment# 53285, Question# 53309, Content# 53301

$\displaystyle\text{Ten quarts}={20}\ \text{pints}$ and $\displaystyle{20}-{6}={14}$

Assessment# 53285, Question# 53309, Content# 53303

$\displaystyle\text{Ten quarts}={20}\ \text{pints}$ , not 10. So after drinking 6 of the 20, 14 are left.

Assessment# 53285, Question# 53309, Content# 53305

You seem to have counted cups instead of pints. $\displaystyle\text{Ten quarts}={40}\ \text{cups}$ , but only 20 pints. If 6 are gone, 14 are left.

Assessment# 53761, Question# 53773, Content# 53765

$\displaystyle{3}\ \text{ tons}\cdot\frac{{{2},{000}\ \text{ pounds}}}{{{1}\ \text{ ton}}}={6},{000}\ \text{ pounds}$ , and $\displaystyle{6},{000}-{1},{000}={5},{000}$ pounds

Assessment# 53285, Question# 53345, Content# 53337

He needs 32 pints and $\displaystyle{32}\ \text{ pints}\cdot\frac{{{1}\ \text{ quart}}}{{{2}\ \text{ pints}}}={16}\ \text{ quarts}$ .

Assessment# 56963, Question# 56997, Content# 57008

You multiplied $\displaystyle\text{A}$ by a factor of 3 instead of 6.

Assessment# 53513, Question# 53515, Content# 53520

$\displaystyle{2}\ \text{ tons}={4},{000}\ \text{ pounds}$ and $\displaystyle{4},{000}\ \text{ pounds}-{1},{750}\ \text{ pounds}={2},{250}\ \text{ pounds}$

Assessment# 53513, Question# 53515, Content# 53524

You may have incorrectly converted 1,750 pounds to tons by moving the decimal point 3 places to the left, and then subtracted 1.75 from 2. You must multiply the number of pounds by $\displaystyle\frac{{{1}\ \text{ ton}}}{{{2},{000}\ \text{ pounds}}}$ if you want to convert from pounds to tons.

Assessment# 53513, Question# 53539, Content# 53615

$\displaystyle{1}\ \text{ ton}={2},{000}\ \text{ pounds}$ and $\displaystyle{2},{500}\ \text{ pounds}+{2},{000}\ \text{ pounds}={4},{500}\ \text{ pounds}$ .

Assessment# 53513, Question# 53599, Content# 53604

Convert the pounds to ounces:

$\displaystyle\frac{{{4}\ \text{ pounds}}}{{{1}}}\cdot\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}={64}\ \text{ ounces}$

Add the two weights: $\displaystyle{64}\ \text{ ounces}+{7}\ \text{ ounces}={71}\ \text{ ounces}$ .

Assessment# 53513, Question# 53599, Content# 53608

You may have multiplied the 4 pounds by $\displaystyle\frac{{{1}\ \text{ pound}}}{{{16}\ \text{ ounces}}}$ before adding it to the 7 ounces. You should multiply by $\displaystyle\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}$ . The correct answer is 71 ounces.

Assessment# 53622, Question# 53634, Content# 53626

$\displaystyle{3}\ \text{ tons}\cdot\frac{{{2},{000}\ \text{ pounds}}}{{{1}\ \text{ ton}}}={6},{000}\ \text{ pounds}$ and $\displaystyle{6},{000}-{2},{400}={3},{600}$ pounds

Assessment# 53622, Question# 53634, Content# 53630

You may have incorrectly converted 2,400 pounds to tons by moving the decimal point 3 places to the left, and then subtracted 2.4 from 3. You must multiply the number of pounds by $\displaystyle\frac{{{1}\ \text{ ton}}}{{{2},{000}\ \text{ pounds}}}$ if you want to convert from pounds to tons. The correct answer is 3,600 pounds.

Assessment# 53622, Question# 53646, Content# 53638

$2\\ " tons="" pounds="$ and $4,000\\ " +1,900\\="" pounds="$

Assessment# 53622, Question# 53672, Content# 53664

$\displaystyle{3}\frac{{{1}}}{{{2}}}=\frac{{{7}}}{{{2}}}$ . Convert the pounds to ounces:

$\displaystyle\frac{{7}}{{2}}\ \text{ pounds}\cdot\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}={56}\ \text{ ounces}$

$\displaystyle{56}+{4}={60}$ ounces

Assessment# 53622, Question# 53672, Content# 53668

You may have multiplied the $\displaystyle{3}\frac{{1}}{{2}}$ pounds by $\displaystyle\frac{{{1}\ \text{ pound}}}{{{16}\ \text{ ounces}}}$ before adding the pounds to the 4 ounces. You should multiply by $\displaystyle\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}$ to get the correct number of ounces in any number of pounds.

Assessment# 52711, Question# 53682, Content# 53691

Convert pounds to ounces:

$\displaystyle{5}\ \text{ pounds}\cdot\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}={80}\ \text{ ounces}$ . Calculate the cost:

$" 0.44="" +="" \\$="" 0.20(79)="\\$" 16.24="$ .

Assessment# 53700, Question# 53712, Content# 53704

$320,000\\ " *(1\\="" pound="" (16\\="" ounces="" )*(1\\="" ton="" (2,000\\="" pounds="" )="" tons="$

Assessment# 53700, Question# 53745, Content# 53737

$56\\ " *(1\\="" pound="" (16\\="" ounces="" )="" pounds="$

Assessment# 53700, Question# 53757, Content# 53749

$32,000\\ " *(1\\="" ton="" (2,000\\="" pounds="" )="" tons="$

Assessment# 53761, Question# 53791, Content# 53783

$\displaystyle{3},{200}+{2},{700}+{1},{100}={7},{000}$ pounds

Convert pounds to tons:

$7,000\\ " *(1\\="" ton="" (2,000\\="" pounds="" )="" tons="$

Assessment# 53761, Question# 53803, Content# 53795

Convert pounds to ounces:

$8\\ " pounds="" *(16\\="" (1\\="" pound="" )="" ounces="$

Calculate the cost:

$" 0.44+="" \\$="" 0.20(127)="\\$" 25.84="$

Assessment# 53761, Question# 53815, Content# 53807

$10,000\\ " *(1\\="" ton="" (2,000\\="" pounds="" )="" tons="$ and $5*" \\$="" 110="\\$" 550="$

Assessment# 53825, Question# 53837, Content# 53829

Converting Bordeaux's temperature to Fahrenheit:

$\displaystyle{F}=\frac{{9}}{{5}}\cdot{25}+{32}$

$\displaystyle{F}=\frac{{225}}{{5}}+{32}={45}+{32}={77}$

$\displaystyle{{95}}^{\circ}\text{F}-{{77}}^{\circ}\text{F}={{18}}^{\circ}\text{F}$

Assessment# 53825, Question# 53837, Content# 53831

You may have mixed up which city had which temperature while calculating. Austin's temperature is $\displaystyle{{95}}^{\circ}\text{F}$ . You need to convert Bordeaux's temperature to Fahrenheit to compare it with Austin's temperature.

Assessment# 53825, Question# 53868, Content# 53860

$\displaystyle{C}=\frac{{5}}{{9}}{\left({F}-{32}\right)}=\frac{{5}}{{9}}{\left({203}-{32}\right)}=\frac{{5}}{{9}}\cdot{171}=\frac{{855}}{{9}}={95}$ or $\displaystyle{{95}}^{\circ}\text{C}$

Assessment# 53825, Question# 53880, Content# 53874

You used the wrong formula. To convert from Celsius to Fahrenheit, use the formula $\displaystyle{F}=\frac{{9}}{{5}}\cdot{C}+{32}$ . The correct answer is $\displaystyle{{99}}^{\circ}\text{F}$ .

Assessment# 53825, Question# 53880, Content# 53876

You used the ratio $\displaystyle\frac{{9}}{{5}}$ in the conversion formula, instead of $\displaystyle\frac{{5}}{{9}}$ . The correct answer is $\displaystyle{{99}}^{\circ}\text{F}$ .

Assessment# 53825, Question# 53880, Content# 53878

You subtracted 32 in the conversion formula, instead of adding 32. The correct answer is $\displaystyle{{99}}^{\circ}\text{F}$ .

Assessment# 53212, Question# 54130, Content# 54135

$\displaystyle{3}\ \text{ pints}\cdot\frac{{{2}\ \text{ cups}}}{{{1}\ \text{ pint}}}\cdot\frac{{{8}\ \text{ fluid ounces}}}{{{1}\ \text{ cup}}}={48}\ \text{ fluid ounces}$ . So there are $\displaystyle{48}+{8}={56}$ fluid ounces.

Assessment# 53285, Question# 53309, Content# 54144

$\displaystyle{10}\ \text{ quarts}\cdot\frac{{{4}\ \text{ cups}}}{{{1}\ \text{ quart}}}\cdot\frac{{{8}\ \text{ fluid ounces}}}{{{1}\ \text{ cup}}}={320}\ \text{ fluid ounces}$

There are $\displaystyle{320}-{48}={272}$ fluid ounces.

Assessment# 54227, Question# 54234, Content# 54245

You may have thought the starting temperature was $\displaystyle-{{7}}^{\circ}\text{F}$ instead of $\displaystyle{{7}}^{\circ}\text{F}$ . The problem asks you to find $\displaystyle{7}-{30}$ .

Assessment# 54227, Question# 54246, Content# 54251

$" 125-="" \\$="" 63.27="\\$" 61.73="$ and $" 61.73-="" 68.10="-" \\$="" 6.37="$ .

Levi is overdrawn, so another $28 must be subtracted. $-" 6.37-="" 28="-" 6.37+(-="" 28)="-" \\$="" 34.37="$

Assessment# 54677, Question# 54679, Content# 54686

You only calculated the change in temperature, and did not add it to the initial temperature of $\displaystyle{{78}}^{\circ}\text{F}$ .

Assessment# 54677, Question# 54691, Content# 54696

$3(-" \\$="" 20)="-"\\$"60"$

Assessment# 54677, Question# 54705, Content# 54710

$3(-" 20)="-" \\$="" 60="$

Assessment# 54677, Question# 54705, Content# 54714

You added $-" \\$="" 20="$ and $-" \\$="" 3="$ instead of multiplying $3(-" \\$="" 20)="$ .

Assessment# 54677, Question# 54705, Content# 54716

You added $-" \\$="" 20="$ and $3 instead of multiplying $3(-" \\$="" 20)="$ .

Assessment# 54677, Question# 54801, Content# 54808

You correctly divided $196 by 7, but each friend's debt is $-" \\$="" 196-:7="$ . A negative number divided by a positive gives a negative quotient, meaning each friend has a debt of $28. The correct answer is $-" \\$="" 28="$ .

Assessment# 54677, Question# 54801, Content# 54810

You multiplied the debt of the group by 7 instead of dividing by 7. The correct answer is $-" \\$="" 28="$ .

Assessment# 54677, Question# 54801, Content# 54812

You mutliplied the debt of the group by 7, instead of dividing by 7, and you failed to give the answer a negative sign. The correct answer is $-" \\$="" 28="$ .

Assessment# 55596, Question# 55632, Content# 55626

You correctly found the difference of the absolute values of the addends. You need to apply the sign of the addend with the greater absolute value. The correct answer is $\displaystyle-{{59}}^{\circ}\text{C}$ .

Assessment# 55596, Question# 55632, Content# 55628

The freezing point for Substance B is found by adding $\displaystyle-{185}+{126}$ . Since the addends have different signs, you must find the difference of the absolute values and give the sum the same sign as the addend with the greater absolute value. The correct answer is $\displaystyle-{{59}}^{\circ}\text{C}$ .

Assessment# 55596, Question# 55632, Content# 55630

The freezing point for Substance B is found by adding $\displaystyle-{185}+{126}$ . You added the absolute values of the addends, and applied a negative sign. Since the addends have different signs, so you must find the difference of the absolute values and give the sum the same sign as the addend with the greater absolute value. The correct answer is $\displaystyle-{{59}}^{\circ}\text{C}$ .

Assessment# 56453, Question# 56507, Content# 56512

You wrote the equation: $4\\" -15="\\" s="" +15="$ and solved it to obtain the right answer.

Assessment# 56106, Question# 56113, Content# 56118

You combined terms in $\displaystyle\text{x}$ and independent terms. Then applied the Multiplicative Property of Equality to isolate the variable and solve.

Assessment# 56106, Question# 56113, Content# 56120

You may have made an error when finding the common denominator of the fractions that are coefficients of $\displaystyle\text{x}$ .

Assessment# 56106, Question# 56113, Content# 56124

It seems like you could have made an error when applying the Multiplicative Property of Equality to the equation just before you solved for $\displaystyle\text{x}$ .

Assessment# 56106, Question# 56125, Content# 56134

You may have made an error when combining terms in $\displaystyle\text{x}$ .

Assessment# 56106, Question# 56125, Content# 56136

You may have made an error when combining terms in $\displaystyle\text{x}$ .

Assessment# 56106, Question# 56137, Content# 56142

By applying the Addition Property of Equality you combined the terms in $\displaystyle\text{y}$ and the independent ones together, then applied the Multiplicative Property of Equality to find the value of $\displaystyle\text{y}$ .

Assessment# 56106, Question# 56137, Content# 56144

You may have applied incorrectly the Addition Property of Equality at the beginning of the solution. If you apply it correctly and then apply the Multiplicative Property of Equality you will arrive at $" y="" \\="9$ as the correct answer.

Assessment# 56106, Question# 56137, Content# 56146

You may have forgotten to apply the Multiplicative Property of Equality to the right side, towards the end of your solution process. If you apply the Addition and Multiplicative Properties of Equality, you will arrive at the correct answer: $" y="" \\="9$ .

Assessment# 56106, Question# 56137, Content# 56148

You may have incorrectly applied the Multiplicative Property of Equality towards the end of your solution. If you correctly apply both the Multiplicative and Addition Properties of Equality you will arrive at $" y="" \\="9$ . the correct answer.

Assessment# 56333, Question# 56385, Content# 56394

If that were the case you would have arrived at $\displaystyle\text{y}={15}$ .

Assessment# 56453, Question# 56467, Content# 56472

Your wrote and solved the correct equation: $2\\" +3="\\" c="" +8="$

Assessment# 56453, Question# 56467, Content# 56474

You must have written and solved: $2\\" -3="\\" c="" +8="$

But that is not the correct equation.

Assessment# 56453, Question# 56467, Content# 56476

At some point in the solution you must have come up incorrectly with $2\\" c="$ .

Assessment# 56453, Question# 56467, Content# 56478

At some point in your solution you must have written, incorrectly: $2\\" c="$

Assessment# 56453, Question# 56489, Content# 56496

That would be a convenient price for a shirt! You must have written the equation: `6\\` and solved to get that answer. However, the correct equation is: $6\\" -20="4\\" s="" +30="$ , and when you solve it you get the right answer: $" s="" \\="\\$" 25="$

Assessment# 56453, Question# 56489, Content# 56498

A shirt at that price would be a real bargain! However, it seems that you wrote the incorrect equation: $6\\" +4\\="" s="$ . and solved it to get that answer. The correct equation is: $6\\$ , and when you solve it you get the right answer: $\displaystyle\text{s}=\text{\$}{25}$

Assessment# 56453, Question# 56489, Content# 56500

A negative price would mean that you get paid to buy the item. In solving the equation you must have come up with $4\\" -6\\="" s="$ , and went on to obtain the negative quantity for an answer. However, the correct equation is: $6\\$ , and when you solve it you get the right answer: $\displaystyle\text{s}=\text{\$}{25}$

Assessment# 56453, Question# 56507, Content# 56514

It seems like you came up with $5\\" s="$ at some point in your solution. Revise your thinking.

Assessment# 56453, Question# 56507, Content# 56516

It seems like you came up with $6\\" s="$ at some point in your solution. Revise your thinking.

Assessment# 56453, Question# 56507, Content# 56518

It looks like you came up with $2\\" s="$ at some point. Revise your thinking.

Assessment# 56963, Question# 56997, Content# 57002

You applied $" a="" \\="(\\" *2)*(l*3)="\\" b="" *l*6="$

Assessment# 56963, Question# 56997, Content# 57004

You multiplied $\displaystyle\text{A}$ by a factor of 5 instead of 6.

Assessment# 56963, Question# 56997, Content# 57006

You multiplied $\displaystyle\text{A}$ by a factor of 2 instead of 6.

Assessment# 57036, Question# 57081, Content# 57091

You seem to have applied the Multiplicative Property of Equality incorrectly. $\displaystyle{l}$ and $\displaystyle\text{w}$ should not be added together.

Assessment# 57036, Question# 57081, Content# 57093

You seem to have subtracted $\displaystyle{l}$ and $\displaystyle\text{w}$ from 315. Apply the Multiplicative Property of Equality to solve correctly.

Assessment# 54227, Question# 59865, Content# 59870

If you start with an amount and subtract the amount you owe, then the result is the amount left over. Beth has $-" \\$="" 2="$ left over, which means she still owes her sister another $2. A negative amount of money means there is a debt.

Assessment# 54227, Question# 59865, Content# 59872

If you start with an amount and subtract the amount you owe, then the result is the amount left over. She has $-" \\$="" 2="$ left over, not $2.

Assessment# 54227, Question# 59865, Content# 59874

The equation is about Beth's money. She has $-" \\$="" 2="$ left over, which means that she has a debt.

Assessment# 60319, Question# 60343, Content# 60335

$" 230-="" \\$="" 127.63="\\$" 102.37="$ and $" 102.37-="" 108.02="-" \\$="" 5.65="$

Tracy is overdrawn, so another $35 must be subtracted. $-" 5.65-="" 35="-" 5.65+(-="" \\$="" 35)="-"\\$"40.65""$

Assessment# 60319, Question# 60353, Content# 60347

If you start with an amount and subtract the amount you owe, then the result is the amount left over. Mark has $-" \\$="" 4="$ left over, which means he still owes his dad another $4. A negative amount of money means there is a debt.

Assessment# 60319, Question# 60353, Content# 60349

If you start with an amount and subtract the amount you owe, then the result is the amount left over. Mark has $-" \\$="" 4="$ left over, not $4.

Assessment# 60319, Question# 60353, Content# 60351

The equation is about Mark's money. He has $-" \\$="" 4="$ left over, which means that he has a debt.

Assessment# 60319, Question# 60365, Content# 60363

You may have thought the starting temperature was $\displaystyle-{{9}}^{\circ}\text{C}$ instead of $\displaystyle{{9}}^{\circ}\text{C}$ . The problem is asking you to find $\displaystyle{9}-{23}$ .

Assessment# 54677, Question# 54801, Content# 60827

You found the debt for the 7 tickets without including the fee. The amount you need to calculate is $\displaystyle{7}{\left(-{28}\right)}+{\left(-{25}\right)}$ . The correct answer is $-" \\$="" 221="$ .

Assessment# 54677, Question# 54801, Content# 60825

You added a $25 fee for each ticket, when the fee was for the entire order. The amount that you need to calculate is $\displaystyle{7}{\left(-{28}\right)}+{\left(-{25}\right)}$ . The correct answer is $-" \\$="" 221="$ .

Assessment# 60904, Question# 60916, Content# 60908

$5(-" 10)="-" \\$="" 50="$

Assessment# 60904, Question# 60916, Content# 60912

You added $-" \\$="" 10="$ and $-" \\$="" 5="$ instead of multiplying $5(-" \\$="" 10)="$ .

Assessment# 60904, Question# 60916, Content# 60914

You added $-" \\$="" 10="$ and $5 instead of multiplying $5(-" \\$="" 10)="$ .

Assessment# 60904, Question# 60940, Content# 60934

You only calculated the change in temperature, and did not add it to the initial temperature of $\displaystyle{{70}}^{\circ}\text{F}$ .

Assessment# 60904, Question# 60952, Content# 60946

You probably correctly found $\displaystyle{9}{\left(-{32}\right)}$ , but then added the $15 to this amount. It is true that the tour company added the fee, but this amount is also a debt, so you should be adding $-" \\$="" 15="$ to the cost of the 9 tickets.

Assessment# 63892, Question# 63933, Content# 63927

You may have divided both sides by 9. That would leave you with $\displaystyle{\left(\frac{{1}}{{27}}\right)}{x}={1}$ . This would not mean that x is equal to $\displaystyle\frac{{1}}{{27}}$ . The correct answer is $" x="" \\="27$ .

Assessment# 63892, Question# 63933, Content# 63929

It seems that you added 3 instead of multiplying both sides of the equation by 3. If you add 3 to $\displaystyle\frac{{x}}{{3}}$ that will not leave you with x, and you must do the same thing to both sides. The correct answer is $" x="" \\="27$ .

Assessment# 63892, Question# 63933, Content# 63931

It seems that you divided by 3 instead of multiplying both sides of the equation by 3. If you divide $\displaystyle\frac{{x}}{{3}}$ by 3 that will not leave you with x, and you must do the same thing to both sides. The correct answer is $" x="" \\="27$ .

Assessment# 72106, Question# 72108, Content# 72113

Rate equals work done over time, or $\displaystyle{r}=\frac{{W}}{{t}}$ . Henry's rate for the previous job was $\displaystyle\frac{{400}}{{2}}$ square feet per hour. His rate for the new job will be $\displaystyle\frac{\text{1,500}}{{t}}$ square feet per hour. These two rates are equal.

Assessment# 72106, Question# 72108, Content# 72115

Use the fact that rate equals work done over time, or $\displaystyle{r}=\frac{{W}}{{t}}$ . Henry's rate for the previous job was 200 square feet per hour. The fraction $\displaystyle\frac{{t}}{\text{1,500}}$ does not express Henry's rate for the new job.

Assessment# 77770, Question# 77782, Content# 77778

You may have chosen this answer because you know that the number 169 has two square roots: 13 and $\displaystyle-{13}$ . However, the symbol $\displaystyle\sqrt{{\ \text{ }\ }}$ means the positive square root.

Assessment# 83091, Question# 83125, Content# 83119

It is true that the symbol $\displaystyle\sqrt{{\ \text{ }\ }}$ stands for the positive square root and that multiplying by a negative will make the square root negative. However, those are facts about the value or output of the function. The domain, however, is the set of all possible values of the input x.

Assessment# 84807, Question# 84831, Content# 84823

First use the power property, then use the quotient property:

$\displaystyle{\log}_{{{10}}}{{\left(\frac{\text{1,000}}{{a}}\right)}}^{{2}}={2}\cdot{\log}_{{{10}}}{\left(\frac{\text{1,000}}{{a}}\right)}={2}{\left({\log}_{{{10}}}\text{1,000}-{\log}_{{{10}}}{a}\right)}$

$\displaystyle={2}{\left({3}\right)}-{2}\cdot{\log}_{{{10}}}{a}={6}-{2}\cdot{\log}_{{{10}}}{a}$

Assessment# 84807, Question# 84831, Content# 84827

You may have incorrectly rewritten the logarithm as $\displaystyle{\log}_{{{10}}}{\text{1,000}}^{{2}}-{\log}_{{10}}{a}$ . You must first use the power property, and then use the quotient property.

Assessment# 84807, Question# 84831, Content# 84829

You may have simplified correctly for the first two steps, but then made a mistake with the factor 2. Start with the power property, then the quotient property: $\displaystyle{\log}_{{{10}}}{{\left(\frac{\text{1,000}}{{a}}\right)}}^{{2}}={2}\cdot{\log}_{{{10}}}{\left(\frac{\text{1,000}}{{a}}\right)}={2}{\left({\log}_{{{10}}}\text{1,000}-{\log}_{{{10}}}{a}\right)}$

If you have a term such as $\displaystyle{2}\cdot{\log}_{{{10}}}{a}$ , you may not rewrite it with the factor 2 inside the logarithm.

Assessment# 99723, Question# 99725, Content# 99746

The side opposite angle A has length 6, the side adjacent to angle A has length 8, and the hypotenuse has length 10.

$\displaystyle{\sin{{A}}}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{{6}}{{10}}$

Assessment# 99723, Question# 99729, Content# 99754

The side opposite angle S has length 3, the side adjacent to angle S has length 2, and the hypotenuse has length $\displaystyle\sqrt{{13}}$ .

$\displaystyle{\cos{{S}}}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{2}}{\sqrt{{13}}}$

$\displaystyle{\tan{{S}}}=\frac{\text{opposite}}{\text{adjacent}}=\frac{{3}}{{2}}$

Assessment# 99723, Question# 99741, Content# 99786

The side opposite angle D has length 9, the side adjacent to angle D has length 40, and the hypotenuse has length 41.

$\displaystyle{\sin{{D}}}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{{9}}{{41}}$

$\displaystyle{\sec{{D}}}=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{{41}}{{40}}$

Assessment# 99795, Question# 99807, Content# 99799

The side opposite angle A has length 6, the side adjacent to angle A has length 8, and the hypotenuse has length 10.

$\displaystyle{\cos{{A}}}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{8}}{{10}}$

Assessment# 99795, Question# 99819, Content# 99811

The side opposite angle N has length $\displaystyle\sqrt{{24}}$ , the side adjacent to angle N has length 5, and the hypotenuse has length 7.

$\displaystyle{\sec{{N}}}=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{{7}}{{5}}$

$\displaystyle{\csc{{N}}}=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{{7}}{\sqrt{{24}}}$

Assessment# 99795, Question# 99843, Content# 99835

The side opposite angle F has length 7, the side adjacent to angle F has length 24, and the hypotenuse has length 25.

$\displaystyle{\cot{{F}}}=\frac{\text{adjacent}}{\text{opposite}}=\frac{{24}}{{7}}$

$\displaystyle{\cos{{F}}}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{24}}{{25}}$

Assessment# 100188, Question# 100195, Content# 100206

It appears that you correctly used the Pythagorean Theorem to find $\displaystyle{h}=\sqrt{{13}}$ , but you found the value of $\displaystyle{\cos{\theta}}$ . Remember the definition is:

$\displaystyle{\sin{\theta}}=\frac{\text{opposite}}{\text{hypotenuse}}$

Assessment# 100188, Question# 100195, Content# 100202

You found the value of $\displaystyle{\tan{\theta}}$ . First use the Pythagorean Theorem to find h:

$\displaystyle{{3}}^{{2}}+{{2}}^{{2}}={{h}}^{{2}}$

After solving for h, use the definition of sine:

$\displaystyle{\sin{\theta}}=\frac{\text{opposite}}{\text{hypotenuse}}$

Assessment# 100188, Question# 100218, Content# 100212

You found the value of $\displaystyle{\sin{{Y}}}$ . First use the Pythagorean Theorem to find x:

$\displaystyle{{5}}^{{2}}+{{x}}^{{2}}={{8}}^{{2}}$

After solving for x, use the definition of tangent:

$\displaystyle\text{tangent}=\frac{\text{opposite}}{\text{adjacent}}$

Assessment# 100188, Question# 100256, Content# 100263

It appears that you correctly used the Pythagorean Theorem to find the length of the hypotenuse. However, you have found the value of $\displaystyle{\sin{{X}}}$ . Remember that:

$\displaystyle\text{cosine}=\frac{\text{adjacent}}{\text{hypotenuse}}$

The correct answer is $\displaystyle\frac{{3}}{\sqrt{{34}}}$ .

Assessment# 100188, Question# 100256, Content# 100267

It appears that you correctly used the Pythagorean Theorem to find the length of the hypotenuse. However, you have found the value of $\displaystyle{\csc{{X}}}$ . Remember that:

$\displaystyle\text{cosine}=\frac{\text{adjacent}}{\text{hypotenuse}}$

The correct answer is $\displaystyle\frac{{3}}{\sqrt{{34}}}$ .

Assessment# 100188, Question# 100279, Content# 100273

You found the value of $\displaystyle{\cos{{B}}}$ . First use the Pythagorean Theorem to find b:

$\displaystyle{{b}}^{{2}}+{{6}}^{{2}}={{7}}^{{2}}$

After solving for b, use the definition of cosecant:

$\displaystyle\text{cosecant}=\frac{\text{hypotenuse}}{\text{opposite}}$

Assessment# 100280, Question# 100292, Content# 100286

You found the value of $\displaystyle{\tan{\theta}}$ . First use the Pythagorean Theorem to find h:

$\displaystyle{{3}}^{{2}}+{{5}}^{{2}}={{h}}^{{2}}$

After solving for h, use the definition of cosine:

$\displaystyle{\cos{\theta}}=\frac{\text{adjacent}}{\text{hypotenuse}}$

Assessment# 100280, Question# 100292, Content# 100290

It appears that you correctly used the Pythagorean Theorem to find $\displaystyle{h}=\sqrt{{34}}$ , but you found the value of $\displaystyle{\sin{\theta}}$ . Remember the definition is:

$\displaystyle{\cos{\theta}}=\frac{\text{adjacent}}{\text{hypotenuse}}$

Assessment# 100280, Question# 100304, Content# 100298

You found the value of $\displaystyle{\sin{{Q}}}$ . Remember that the two acute angles in a right triangle will have different trigonometric ratios. First use the Pythagorean Theorem to find p:

$\displaystyle{{7}}^{{2}}+{{p}}^{{2}}={{10}}^{{2}}$

After solving for p, use the definition of sine:

$\displaystyle\text{sine}=\frac{\text{opposite}}{\text{hypotenuse}}$

Assessment# 100280, Question# 100304, Content# 100302

It appears that you correctly used the Pythagorean Theorem to find $\displaystyle{p}=\sqrt{{51}}$ , but you found the value of $\displaystyle{\tan{{P}}}$ . Use the definition of sine:

$\displaystyle\text{sine}=\frac{\text{opposite}}{\text{hypotenuse}}$

Assessment# 100280, Question# 100316, Content# 100310

It appears that you correctly used the Pythagorean Theorem to find the length of the hypotenuse. However, you have found the value of $\displaystyle{\csc{{X}}}$ . Remember that:

$\displaystyle\text{secant}=\frac{\text{hypotenuse}}{\text{adjacent}}$

Assessment# 100280, Question# 100316, Content# 100314

It appears that you correctly used the Pythagorean Theorem to find the length of the hypotenuse. However, you have found the value of $\displaystyle{\sin{{X}}}$ . Remember that:

$\displaystyle\text{secant}=\frac{\text{hypotenuse}}{\text{adjacent}}$

Assessment# 100280, Question# 100328, Content# 100322

You found the value of $\displaystyle{\cos{{E}}}$ . First use the Pythagorean Theorem to find e:

$\displaystyle{{e}}^{{2}}+{{4}}^{{2}}={{9}}^{{2}}$

After solving for e, use the definition of cotangent:

$\displaystyle\text{cotangent}=\frac{\text{adjacent}}{\text{hypotenuse}}$

The correct answer is $\displaystyle\frac{{4}}{\sqrt{{65}}}$ .

Assessment# 100343, Question# 100345, Content# 100350

Use the definition of tangent together with the given value:

$\displaystyle{{\tan{{15}}}}^{\circ}=\frac{\text{opposite}}{\text{adjacent}}$

$\displaystyle{0.27}\approx\frac{{x}}{{7}}$

$\displaystyle{7}{\left({0.27}\right)}\approx{x}$

$\displaystyle{1.89}\approx{x}$ , which rounds off to 1.9.

Assessment# 100392, Question# 100404, Content# 100396

Use the definition of cosine together with the given value:

$\displaystyle{{\cos{{20}}}}^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\displaystyle{0.94}\approx\frac{{m}}{{6}}$

$\displaystyle{6}{\left({0.94}\right)}\approx{m}$

$\displaystyle{5.64}\approx{x}$ , which rounds off to 5.6.

Assessment# 101350, Question# 101405, Content# 101410

To use the formula for radian measure, you must first have the radius and the arc length in the same units, so you must convert one of them. Now

$\displaystyle{\left(\frac{{{2}\pi}}{{9}}\ \text{ yards}\right)}\cdot{\left(\frac{{{3}\ \text{ feet}}}{{{1}\ \text{ yard}}}\right)}=\frac{{{2}\pi}}{{3}}\ \text{ feet}$ , so:

$\displaystyle\theta=\frac{{s}}{{r}}=\frac{{\frac{{{2}\pi}}{{3}}\ \text{ ft}}}{{{2}\ \text{ ft}}}=\frac{{{2}\pi}}{{3}}\cdot\frac{{1}}{{2}}=\frac{\pi}{{3}}$

Assessment# 101417, Question# 101462, Content# 101454

To use the formula for radian measure, you must first have the radius and the arc length in the same units, so you must convert one of them.

Now: $\displaystyle{\left(\frac{{{3}\pi}}{{4}}\ \text{ feet}\right)}{\left(\frac{{{1}\ \text{ yard}}}{{{3}\ \text{ feet}}}\right)}=\frac{\pi}{{4}}\ \text{ yards}$

So: $\displaystyle\theta=\frac{{s}}{{r}}=\frac{{\frac{\pi}{{4}}\ \text{ yd}}}{{\frac{{1}}{{2}}\ \text{ yd}}}=\frac{\pi}{{4}}\cdot\frac{{2}}{{1}}=\frac{\pi}{{2}}$

Assessment# 102290, Question# 102302, Content# 102296

The definition of amplitude is $\displaystyle\frac{{\text{maximum}-\text{minimum}}}{{2}}$ . The expression $\displaystyle{\left|{a}\right|}$ will give you the same value.

Assessment# 102290, Question# 102302, Content# 102300

The definition of amplitude is $\displaystyle\frac{{\text{maximum}-\text{minimum}}}{{2}}$ . This turns out to be equal to the height of one hill, or the depth of one valley.

Assessment# 118115, Question# 129773, Content# 129769

You may have multiplied 42 inches by $(12\\ " inches="" (1\\="" foot="" )="$ . You need to multiply 42 inches by $(1\\ " foot="" (12\\="" inches="" )="$ to get the correct answer.

Assessment# 117871, Question# 117883, Content# 117875

The Fresh Water rate is $" \\$1.76"="" "16="" ounces"="\\$0.11" "1="" ounce""="$

and the Spring Water rate is $" \\$2.40"="" "20="" ounces"="\\$0.12" "1="" ounce""="$

Assessment# 117871, Question# 117895, Content# 117887

The Kitty Kibbles rate is $" \\$24.00"="" "16="" pounds"="\\$1.50" "1="" pound""="$

and the Feline Flavor rate is $(" 11.20)="" (8="" pounds="" pound="" )="(" \\$="" 1.40)="" (1="$ .

Assessment# 117961, Question# 117973, Content# 117965

$\displaystyle\frac{{n}}{\text{35 attempts}}=\frac{\text{4 made}}{\text{5 attempts}}$

$\displaystyle{n}\cdot{5}={35}\cdot{4}$

$\displaystyle{n}=\frac{{140}}{{5}}$

$\displaystyle{n}={28}$

Assessment# 118115, Question# 129761, Content# 129757

You may have multiplied 44 inches by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ . You need to multiply 44 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 118145, Question# 129785, Content# 129779

You may have multiplied 3 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 8 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 5 inches.

Assessment# 118115, Question# 118127, Content# 118121

Assessment# 118115, Question# 118127, Content# 118123

You may have multiplied 72 inches by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ . You need to multiply 72 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 118115, Question# 118127, Content# 118125

You may have multiplied 72 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{10}\ \text{ inches}}}$ . You need to multiply 72 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 118145, Question# 129785, Content# 129777

$\displaystyle{3}\ \text{ feet}\cdot\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}={36}\ \text{ inches}$ and $\displaystyle\text{36 inches + 8 inches = 44 inches}$ .

Assessment# 118115, Question# 118139, Content# 118133

Assessment# 118115, Question# 118139, Content# 118135

You may have multiplied 78 inches by $(12\\ " inches="" (1\\="" foot="" )="$ . You need to multiply 78 inches by $(1\\ " foot="" (12\\="" inches="" )="$ to get the correct answer.

Assessment# 118115, Question# 118139, Content# 118137

You may have multiplied 78 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{10}\ \text{ inches}}}$ . You need to multiply 78 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 118145, Question# 118157, Content# 118149

$\displaystyle\text{4 feet}\cdot\frac{\text{12 inches}}{\text{1 foot}}=\text{48 inches}$ and $\displaystyle\text{48 inches + 5 inches = 53 inches}$ .

Assessment# 118145, Question# 118157, Content# 118151

You may have multiplied 4 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 5 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 5 inches.

Assessment# 118145, Question# 118157, Content# 118153

You may have multiplied 4 feet by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ and then added 5 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 5 inches.

Assessment# 118145, Question# 118169, Content# 118161

$4\\ " feet="" *(12\\="" (1\\="" foot="" )="" inches="$ and $\displaystyle\text{48 inches + 6 inches = 54 inches}$

Assessment# 118145, Question# 118169, Content# 118163

You may have multiplied 4 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 6 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 6 inches.

Assessment# 118145, Question# 118169, Content# 118165

You may have multiplied 4 feet by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ and then added 6 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 6 inches.

Assessment# 118175, Question# 118187, Content# 118179

$\displaystyle\text{300 yard}\cdot\frac{\text{3 feet}}{\text{1 yard}}=\text{900 feet}$ . To find the perimeter of a rectangle, you can use the formula $\displaystyle\text{perimeter = length + length + width + width}$ .

$\displaystyle\text{900 feet + 900 feet + 200 feet + 200 feet = 2,200 feet}$

Assessment# 118175, Question# 118199, Content# 118191

$\displaystyle\text{130 yards}\cdot\frac{\text{3 feet}}{\text{1 yard}}=\text{390 feet}$

To find the perimeter of a rectangle, you can use the formula $\displaystyle\text{perimeter}=\text{length}+\text{length}+\text{width}+\text{width}$ .

$\displaystyle\text{390 feet + 390 feet + 200 feet + 200 feet}=\text{1,180 feet}$

Assessment# 118175, Question# 118199, Content# 118193

You may have converted the 200 feet to $\displaystyle{66}\frac{{2}}{{3}}\ \text{ yards}$ , then found the total perimeter in yards. You need to convert the 130 yards to feet, then find the total perimeter.

Assessment# 118235, Question# 118247, Content# 118239

$\displaystyle\text{1 ton = 2,000 pounds}$

so $\displaystyle\text{2 tons = 4,000 pounds}$

and $\displaystyle\text{2,500 pounds + 4,000 pounds = 6,500 pounds}$ .

Assessment# 118235, Question# 118259, Content# 118251

$\displaystyle{3}\frac{{{1}}}{{{2}}}=\frac{{{7}}}{{{2}}}$ . Convert the pounds to ounces:

$\displaystyle\frac{{7}}{{2}}\ \text{ pounds}\cdot\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}={56}\ \text{ ounces}$

$\displaystyle{56}+{2}={58}\text{}$ ounces

Assessment# 118235, Question# 118259, Content# 118255

You may have multiplied the $\displaystyle{3}\frac{{1}}{{2}}$ pounds by $\displaystyle\frac{{{1}\ \text{ pound}}}{{{16}\ \text{ ounces}}}$ before adding the pounds to the 2 ounces. You should multiply by $\displaystyle\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}$ to get the correct number of ounces in any number of pounds.

Assessment# 118265, Question# 118277, Content# 118269

$\displaystyle\text{3,200 pounds + 9,200 pounds + 600 pounds = 13,000 pounds}$

Convert pounds to tons:

$\displaystyle\text{13,000 pounds}\cdot\frac{\text{1 ton}}{\text{2,000 pounds}}={6}\frac{{1}}{{2}}\ \text{ tons}$

Assessment# 118265, Question# 118289, Content# 118281

Convert pounds to ounces:

$\displaystyle\text{5 pounds}\cdot\frac{\text{16 ounces}}{\text{1 pound}}=\text{80 ounces}$

Calculate the cost:

$" \\$0.44="" +="" \\$0.20(79)="\\$16.24""$

Assessment# 118295, Question# 118307, Content# 118299

$\displaystyle\text{3 pints}\cdot\frac{\text{2 cups}}{\text{1 pint}}\cdot\frac{\text{8 ounces}}{\text{1 cup}}=\text{48 ounces}$

Assessment# 118325, Question# 118349, Content# 118341

Assessment# 118355, Question# 129953, Content# 129945

He needs 40 pints.

$\displaystyle\frac{{{3}\ \text{ gallons}}}{{{1}}}\cdot\frac{{{4}\ \text{ quarts}}}{{{1}\ \text{ gallon}}}\cdot\frac{{{2}\ \text{ pints}}}{{{1}\ \text{ quart}}}={24}\ \text{ pints}$ , while

$\displaystyle\frac{{{2}\ \text{ gallons}}}{{{1}}}\cdot\frac{{{4}\ \text{ quarts}}}{{{1}\ \text{ gallon}}}\cdot\frac{{{2}\ \text{ pints}}}{{{1}\ \text{ quart}}}={16}\ \text{ pints}$ .

Assessment# 118355, Question# 118367, Content# 118359

He needs 16 pints.

$\displaystyle\frac{{{3}\ \text{ gallons}}}{{{1}}}\cdot\frac{{{4}\ \text{ quarts}}}{{{1}\ \text{ gallon}}}\cdot\frac{{{2}\ \text{ pints}}}{{{1}\ \text{ quart}}}={24}\ \text{ pints}$ , while

$\displaystyle\frac{{{2}\ \text{ gallons}}}{{{1}}}\cdot\frac{{{4}\ \text{ quarts}}}{{{1}\ \text{ gallon}}}\cdot\frac{{{2}\ \text{ pints}}}{{{1}\ \text{ quart}}}={16}\ \text{ pints}$ .

Assessment# 118475, Question# 118487, Content# 118485

$\displaystyle\frac{\text{747 grams}}{\text{1}}\cdot\frac{\text{1 kilogram}}{\text{1,000 grams}}=\text{0.747 kilogram}$

$\displaystyle\text{0.786 kg - 0.747 kg = 0.039 kg}$

Assessment# 118475, Question# 118499, Content# 118491

The distance between the cities is $\displaystyle\frac{\text{920,000 meters}}{\text{1}}\cdot\frac{\text{1 kilometer}}{\text{1,000 meters}}$ or 920 kilometers.

$\displaystyle\frac{\text{920 km}}{{n}}=\frac{\text{230 km}}{\text{1 hour}}$

$\displaystyle{n}=\frac{\text{920}}{{230}}\text{}={4}$

Assessment# 118475, Question# 118499, Content# 118495

You may have correctly converted the distance between cities to 920 kilometers, and then solved the proportion $\displaystyle\frac{{230}}{{n}}=\frac{{920}}{{1}}$ . The correct equation would be

$\displaystyle\frac{\text{920 km}}{{n}}=\frac{\text{230 km}}{\text{1 hr}}$ .

Assessment# 118535, Question# 130111, Content# 130103

$\displaystyle{F}=\frac{\text{9}}{\text{5}}\cdot{C}+{32}=\frac{\text{9}}{\text{5}}\cdot{50}+{32}={90}+{32}={122}$ or $\displaystyle{{122}}^{\circ}\text{F}$

Assessment# 118505, Question# 118529, Content# 118523

Assessment# 118505, Question# 118529, Content# 118527

Assessment# 118535, Question# 118547, Content# 118539

`F="9"/"5"*C + 32 = "9"/"5"*24 + 32 = "216"/"5" + 32 = 43 "1"/"5" + 32 = 75 "1"/"5" ` or $\displaystyle{{75}}^{\circ}\text{F}$

Assessment# 118535, Question# 118559, Content# 118551

$\displaystyle{C}=\frac{\text{5}}{\text{9}}\cdot\text{(}{F}\ \text{ - 32)}=\frac{\text{5}}{\text{9}}\cdot\text{(230 - 32)}=\frac{\text{5}}{\text{9}}\cdot\text{198}=\frac{\text{990}}{\text{9}}=\text{110}$ or $\displaystyle{{110}}^{\circ}\text{C}$

Assessment# 119409, Question# 119421, Content# 119413

To find the mean, add the scores and divide that sum by 5, the number of scores:

$\displaystyle\frac{\text{88 + 81 + 93 +85 +98}}{\text{5}}=\frac{{445}}{{5}}={89}$

To find the median, arrange the values in order from least to greatest and identify the middle value: 81, 85, 88, 93, 98. The median is 88.

The mean with a value of 89 is larger than the median whose value is 88.

Assessment# 120315, Question# 120339, Content# 120333

You probably correctly found $\displaystyle{8}{\left(-{32}\right)}$ , but then added the $10 to this amount. It is true that the tour company added the fee, but this amount is also a debt, so you should be adding $-" \\$10""="$ to the cost of the 8 tickets.

Assessment# 126095, Question# 126107, Content# 126099

The side opposite angle D has length 9, the side adjacent to angle D has length 40, and the hypotenuse has length 41.

$\displaystyle{\sin{{D}}}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{{9}}{{41}}$

$\displaystyle{\cos{{D}}}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{40}}{{41}}$

Assessment# 126215, Question# 126239, Content# 126233

It appears that you correctly used the Pythagorean Theorem to find the length of the hypotenuse. However, you have found the value of $\displaystyle{\csc{{X}}}$ . Remember that:

$\displaystyle\text{secant}=\frac{\text{hypotenuse}}{\text{adjacent}}$

Assessment# 118115, Question# 129773, Content# 129771

Assessment# 117841, Question# 129440, Content# 129432

The rate is $\displaystyle\frac{\text{511 bottles}}{\text{7 minutes}}$ . Express the fraction with 1 in the denominator to find the number of bottles per minute.

Assessment# 118145, Question# 129785, Content# 129781

Assessment# 117871, Question# 129461, Content# 129453

The Spring Water rate is $" \\$2.08"="" "16="" ounces"="\\$0.13" "1="" ounce""="$

and the Fresh Water rate is $(" 2.40)="" (20="" ounces="" ounce="" )="(" \\$="" 0.12)="" (1="$ .

Assessment# 117871, Question# 129473, Content# 129465

The Feline Flavor rate is $(" 20.80)="" (16="" pounds="" pound="" )="(" \\$="" 1.30)="" (1="$

and the Kitty Kibbles rate is $(" 11.20)="" (8="" pounds="" pound="" )="(" \\$="" 1.40)="" (1="$ .

Assessment# 118115, Question# 129761, Content# 129759

You may have multiplied 44 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{10}\ \text{ inches}}}$ . You need to multiply 44 inches by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ to get the correct answer.

Assessment# 118115, Question# 129761, Content# 129755

Assessment# 118115, Question# 129773, Content# 129767

Assessment# 118145, Question# 129797, Content# 129791

You may have multiplied 5 feet by $\displaystyle\frac{{{10}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then added 3 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 3 inches.

Assessment# 118145, Question# 129797, Content# 129793

You may have multiplied 5 feet by $\displaystyle\frac{{{1}\ \text{ foot}}}{{{12}\ \text{ inches}}}$ and then added 3 inches. You need to multiply by $\displaystyle\frac{{{12}\ \text{ inches}}}{{{1}\ \text{ foot}}}$ and then add 3 inches.

Assessment# 118175, Question# 129821, Content# 129813

$\displaystyle\text{230 yards}\cdot\frac{\text{3 feet}}{\text{1 yard}}=\text{690 feet}$

To find the perimeter of a rectangle, you can use the formula $\displaystyle\text{perimeter}=\text{length}+\text{length}+\text{width}+\text{width}$ .

$\displaystyle\text{690 feet + 690 feet + 200 feet + 200 feet}=\text{1,780 feet}$

Assessment# 118175, Question# 129821, Content# 129815

You may have converted the 200 feet to $\displaystyle{66}\frac{{2}}{{3}}\ \text{ yards}$ , then found the total perimeter in yards. You need to convert the 230 yards to feet, then find the total perimeter.

Assessment# 118235, Question# 129869, Content# 129861

$\displaystyle{3}\frac{{{1}}}{{{2}}}=\frac{{{7}}}{{{2}}}$ . Convert the pounds to ounces:

$\displaystyle\frac{{7}}{{2}}\ \text{ pounds}\cdot\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}={56}\ \text{ ounces}$

$\displaystyle{56}+{6}={62}$ ounces

Assessment# 118235, Question# 129869, Content# 129865

You may have multiplied the $\displaystyle{3}\frac{{1}}{{2}}$ pounds by $\displaystyle\frac{{{1}\ \text{ pound}}}{{{16}\ \text{ ounces}}}$ before adding the pounds to the 6 ounces. You should multiply by $\displaystyle\frac{{{16}\ \text{ ounces}}}{{{1}\ \text{ pound}}}$ to get the correct number of ounces in any number of pounds.

Assessment# 118265, Question# 129881, Content# 129873

$\displaystyle\text{3,200 pounds + 7,200 pounds + 600 pounds = 11,000 pounds}$

Convert pounds to tons:

$\displaystyle\text{11,000 pounds}\cdot\frac{\text{1 ton}}{\text{2,000 pounds}}={5}\frac{{1}}{{2}}\ \text{ tons}$

Assessment# 118325, Question# 129941, Content# 129933

Assessment# 118475, Question# 130057, Content# 130055

$\displaystyle\frac{\text{0.747 kilogram}}{\text{1}}\cdot\frac{\text{1,000 grams}}{\text{1 kilogram}}=\text{747 grams}$

$\displaystyle\text{776 grams - 747 grams = 29 grams}$

Assessment# 118475, Question# 130069, Content# 130061

The distance between the cities is $\displaystyle\frac{\text{900,000 meters}}{\text{1}}\cdot\frac{\text{1 kilometer}}{\text{1,000 meters}}$ , or 900 kilometers.

$\displaystyle\frac{\text{900 km}}{{n}}=\frac{\text{180 km}}{\text{1 hour}}$

$\displaystyle{n}=\frac{\text{900}}{{180}}\ \text{ = }\ {5}\text{}$

Assessment# 118475, Question# 130069, Content# 130065

You may have correctly converted the distance between cities to 900 kilometers, and then solved the proportion $\displaystyle\frac{{180}}{{n}}=\frac{{900}}{{1}}$ . The correct equation would be $\displaystyle\frac{\text{900 km}}{{n}}=\frac{\text{180 km}}{\text{1 hr}}$ .

Assessment# 118475, Question# 130069, Content# 130067

You may not have converted correctly from meters to kilometers.

$\displaystyle\frac{\text{900,000 meters}}{\text{1}}\cdot\frac{\text{1 kilometer}}{\text{1,000 meters}}=\text{900 km}$

Assessment# 118505, Question# 130099, Content# 130093

Assessment# 118505, Question# 130099, Content# 130097

Assessment# 118535, Question# 130123, Content# 130115

$\displaystyle{C}=\frac{\text{5}}{\text{9}}\cdot\text{(}{F}\ \text{ - 32)}=\frac{\text{5}}{\text{9}}\cdot\text{(302 - 32)}=\frac{\text{5}}{\text{9}}\cdot\text{270}=\frac{\text{1350}}{\text{9}}=\text{150}$ or $\displaystyle{{150}}^{\circ}\text{C}$

Assessment# 120315, Question# 131386, Content# 131380

You probably correctly found $\displaystyle{12}{\left(-{32}\right)}$ , but then added the $20 to this amount. It is true that the tour company added the fee, but this amount is also a debt, so you should be adding $-" \\$20""="$ to the cost of the 12 tickets.

Assessment# 126095, Question# 140531, Content# 140523

The side opposite angle D has length 9, the side adjacent to angle D has length 40, and the hypotenuse has length 41.

$\displaystyle{\tan{{D}}}=\frac{\text{opposite}}{\text{adjacent}}=\frac{{9}}{{40}}$

$\displaystyle{\sec{{D}}}=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{{41}}{{40}}$

Assessment# 126215, Question# 140679, Content# 140673

It appears that you correctly used the Pythagorean Theorem to find the length of the adjacent side. However, you have found the value of $\displaystyle{\csc{{X}}}$ . Remember that:

$\displaystyle\text{secant}=\frac{\text{hypotenuse}}{\text{adjacent}}$

Assessment# 126215, Question# 140679, Content# 140677

It appears that you correctly used the Pythagorean Theorem to find the length of the adjacent side. However, you have found the value of $\displaystyle{\sin{{X}}}$ . Remember that:

$\displaystyle\text{secant}=\frac{\text{hypotenuse}}{\text{adjacent}}$

Assessment# 126455, Question# 140835, Content# 140827

To use the formula for radian measure, you must first have the radius and the arc length in the same units, so you must convert one of them. Now

$\displaystyle{\left(\frac{{\pi}}{{6}}\ \text{ ft}\right)}\cdot{\left(\frac{{\text{12 in}}}{\text{1 ft}}\right)}={2}\pi\ \text{ in}$ , so:

$\displaystyle\theta=\frac{{s}}{{r}}=\frac{{{2}\pi\ \text{ in}}}{{{12}\ \text{ in}}}=\frac{\pi}{{6}}$