Quantitative Reasoning

Use the conversion fact \(1\text{ m} = 100\text{ cm}\).

Since centimeters are smaller than meters, the number gets larger by a factor of \(100\): \(8 \times 100 = 800\).

So \(8\text{ m} = 800\text{ cm}\).

Use the conversion fact \(1\text{ gal} = 16\text{ cups}\).

Rewrite \(4\tfrac{1}{4}\) gallons as \(4.25\) gallons, then multiply: \(4.25 \times 16 = 68\).

So \(4\tfrac{1}{4}\) gallons is \(68\) cups.

Use the conversion fact \(1\text{ L} = 1000\text{ mL}\).

Set up the dimensional-analysis chain: \(193\text{ tubs} \times 165\text{ mL per tub} = 31{,}845\text{ mL}\).

Then convert milliliters to liters: \(31{,}845 \div 1000 = 31.845\text{ L}\).

Use the temperature formula \(C = \frac{5}{9}(F - 32)\).

Substitute \(F = 41\): \(C = \frac{5}{9}(41 - 32) = \frac{5}{9}(9) = 5\).

Rounded to the nearest tenth, the temperature is \(5.0^\circ\text{C}\).

Use the metric conversion fact \(100\text{ cm} = 1\text{ m}\).

So convert centimeters to meters by dividing by \(100\): \(339 \div 100 = 3.39\).

That gives \(339\text{ cm} = 3.39\text{ m}\).

Use \(1\text{ lb} = 16\text{ oz}\) and \(1\text{ ton} = 2000\text{ lb}\), so \(1\text{ ton} = 32{,}000\text{ oz}\).

Convert ounces to tons: \(456 \div 32{,}000 = 0.01425\).

Rounded to three decimal places, \(456\text{ oz} \approx 0.014\text{ ton}\).

First use the left triangle. Its three angles are \(63^\circ\), \(66^\circ\), and the lower angle at the intersection point.

That lower angle is \(180 - 63 - 66 = 51^\circ\). Since that angle and \(y\) form a straight line, \(y = 180 - 51 = 129^\circ\).

Now use the top triangle: \(28 + 129 + x = 180\), so \(x = 23^\circ\).

Because \(\angle COD\) is a right angle, \(\angle AOC\) is also \(90^\circ\). Since \(\angle BOC = 28^\circ\), the complementary angle is \(\angle AOB = 90 - 28 = 62^\circ\).

Supplementary angles add to \(180^\circ\), so the angle supplementary to \(\angle BOC\) is \(\angle COE\). Vertical angles also show that \(m\angle EOF = 28^\circ\).

Then \(m\angle AOE = 90 + 28 = 118^\circ\) and \(m\angle BOF = 180 - 28 = 152^\circ\).

Use the Pythagorean Theorem: \(x^2 + 8^2 = 13^2\).

That gives \(x^2 + 64 = 169\), so \(x^2 = 105\).

Therefore \(x = \sqrt{105}\). This radical does not simplify any further, so \(\sqrt{105}\) is the exact answer.

Apply the Pythagorean Theorem: \(10^2 + 11^2 = c^2\).

That gives \(100 + 121 = c^2\), so \(221 = c^2\).

Take the square root to get \(c = \sqrt{221}\). Since \(221 = 13 \cdot 17\), the radical does not simplify further.

To get the perimeter, first find the missing inside lengths. The horizontal notch is \(8 - 4 = 4\), and the vertical notch is \(10 - 3 = 7\).

Now add all six outside sides: \(4 + 7 + 4 + 3 + 8 + 10 = 36\). So the perimeter is \(36\) units.

Break the L-shape into two rectangles. The tall rectangle has area \(4 \times 11 = 44\text{ m}^2\).

The short bottom-right rectangle has width \(8 - 4 = 4\) meters and height \(2\) meters, so its area is \(4 \times 2 = 8\text{ m}^2\).

Add them together: \(44 + 8 = 52\text{ m}^2\).

A perimeter problem uses the outside boundary only, so add the four side lengths.

\(21.9 + 10.6 + 23.7 + 26.5 = 82.7\). The height is extra information for area, not perimeter.

Use the trapezoid area formula: \(A = \tfrac{1}{2}(b_1 + b_2)h\).

Substitute the values: \(A = \tfrac{1}{2}(9.8 + 24.5)(19.6) = \tfrac{1}{2}(34.3)(19.6) = 336.14\).

The shaded region is the area of the large circle minus the area of the small circle.

Using \(3.14\) for \(\pi\), \(A = 3.14(9^2) - 3.14(6^2) = 3.14(81 - 36) = 3.14(45) = 141.3\) square units.

For a sphere, use \(V = \tfrac{4}{3}\pi r^3\). The radius is half the diameter, so \(r = 5.5\) cm.

Using \(\pi \approx 3.14\), \(V = \tfrac{4}{3}(3.14)(5.5)^3 \approx 696.56\text{ cm}^3\). Round to two decimal places.

First find how much water Janine drinks in one year: \(6(16.9) = 101.4\) ounces per day, so \(101.4 \times 365 = 37{,}011\) ounces per year. Since \(128\) ounces is \(1\) gallon, she drinks \(37{,}011 \div 128 \approx 289.15\) gallons per year.

Bottled water costs \(\$3.39 \div 24 = \$0.14125\) per bottle. At 6 bottles per day, the yearly bottled-water cost is \(6 \times 365 \times 0.14125 = \$309.34\).

Now compare the filter options, including the reusable bottle. Faucet-mounted: \(\$28 + \$10 + (289.15 - 100) \div 100 \times 11 \approx \$58.81\). Pitcher: \(\$22 + \$10 + (289.15 - 40) \div 40 \times 5 \approx \$63.14\). Under-sink: \(\$130 + \$10 = \$140.00\) because the included 500-gallon filter lasts the whole year.

The faucet-mounted filter is the cheapest option. It saves about \(\$309.34 - \$58.81 = \$250.53\) over one year compared with bottled water.

Convert the speed to feet per second: \(750\text{ mi/hr} \times \frac{5280\text{ ft}}{1\text{ mi}} \times \frac{1\text{ hr}}{3600\text{ s}} = 1100\text{ ft/s}\).

The echo time is a round trip, so in \(2.5\) seconds the sound travels \(1100 \times 2.5 = 2750\) feet total. The canyon wall is half that distance away: \(2750 \div 2 = 1375\) feet.

If the echo takes \(n\) seconds, then the sound travels \(1100n\) feet round trip, so the one-way distance to the wall is \(\frac{1100n}{2} = 550n\) feet.

First find the total number of pages: \(103 \times 12 = 1236\) pages.

A ream has \(500\) pages and costs \(\$3.80\), so the cost per page is \(\$3.80 \div 500 = \$0.0076\).

Now multiply by the number of pages used: \(1236 \times 0.0076 = 9.3936\). Rounded to the nearest cent, the paper cost is \(\$9.39\).

Each pie needs \(1\tfrac{1}{4} = 1.25\) cups of pecans, so for 8 pies you need \(8 \times 1.25 = 10\) cups.

The nutrition label says \(1\) cup weighs \(99\) grams, so \(10\) cups weighs \(10 \times 99 = 990\) grams.

Since \(1\) pound is about \(453.6\) grams, \(990 \div 453.6 \approx 2.18\) pounds. To at least one decimal place, you should buy \(2.2\) pounds of pecans.

Each meatloaf needs \(1\tfrac{1}{4} = 1.25\) cups of breadcrumbs, so for 5 meatloafs you need \(5 \times 1.25 = 6.25\) cups.

Each canister has about \(14\) servings of \(\tfrac{1}{3}\) cup, so one canister holds \(14 \times \tfrac{1}{3} = \tfrac{14}{3} \approx 4.67\) cups.

Now divide the total breadcrumbs needed by the amount in one canister: \(6.25 \div 4.67 \approx 1.34\). To at least one decimal place, that is \(1.3\) canisters, so in practice you would need to buy 2 whole canisters.

Pizza calories are best estimated by area, and the area of a circle depends on the square of the radius. A 6-inch pizza has radius \(3\), and a 16-inch pizza has radius \(8\).

So the area scale factor is \(\frac{8^2}{3^2} = \frac{64}{9}\). Multiply the calories by that factor to estimate the full 16-inch pizza: \(630 \times \frac{64}{9} = 4480\) calories.

The large pizza is cut into 8 slices, so one slice has about \(4480 \div 8 = 560\) calories.

The dosage is given in milligrams per kilogram, so first convert the child's weight from pounds to kilograms: \(29 \div 2.2 \approx 13.18\text{ kg}\).

Now apply the dosage rule: \(13.18 \times 5 \approx 65.9\text{ mg}\). Rounded to the nearest milligram, the usual dose is \(66\text{ mg}\).

First compare the chicken prices. The close store costs \(6 \times \$3.29 = \$19.74\), and the farther store costs \(6 \times \$3.19 = \$19.14\).

Now include gas for the round trip. Close store: \(2(2.1) = 4.2\) miles, so gas costs \(4.2 \div 22 \times \$3.59 \approx \$0.69\). Farther store: \(2(8.5) = 17\) miles, so gas costs \(17 \div 22 \times \$3.59 \approx \$2.78\).

Total cost is about \(\$19.74 + \$0.69 = \$20.43\) for the close store and \(\$19.14 + \$2.78 = \$21.92\) for the farther store. Going to the close store is cheaper, and it saves about \(\$21.92 - \$20.43 = \$1.49\).

The current 80% average applies to the 75% of the course that is already complete, so that part contributes \(0.75(80) = 60\) points toward the final course grade.

If your friend earns 100% on the final, the best possible course grade is \(60 + 0.25(100) = 85\), so the highest overall grade is \(85.0\%\).

For a 75% course grade, solve \(60 + 0.25x = 75\). Then \(0.25x = 15\), so \(x = 60\). The minimum final-exam score needed is \(60.0\%\).

The fill time is proportional to volume. The smaller aquarium has volume \(8 \times 9 \times 11 = 792\) cubic inches.

The larger aquarium has volume \(22 \times 27 \times 31 = 18{,}414\) cubic inches, so it is \(18{,}414 \div 792 = 23.25\) times as large.

Multiply the time by the same factor: \(2 \times 23.25 = 46.5\) minutes. Rounded to the nearest minute, it will take \(47\) minutes.

Each pie needs \(\tfrac{1}{2}\) cup of juice, so two pies need \(1\) cup total. Since \(1\) cup is \(16\) tablespoons, you need \(16\) tablespoons of lemon juice.

Each lemon gives about \(2\) tablespoons, so the number of lemons needed is \(16 \div 2 = 8\).

At \(\$0.54\) per lemon, the total cost is \(8 \times \$0.54 = \$4.32\).

The difference in sales is \(\$330{,}000 - \$230{,}000 = \$100{,}000\).

For part a, compare that difference to Portland's sales: \(100{,}000 \div 230{,}000 \approx 0.4348\), so Seattle's sales were \(43.5\%\) larger than Portland's.

For part b, compare the same difference to Seattle's sales: \(100{,}000 \div 330{,}000 \approx 0.3030\), so Portland's sales were \(30.3\%\) smaller than Seattle's.

For part c, compare Portland directly to Seattle: \(230{,}000 \div 330{,}000 \approx 0.6970\), so Portland's sales were \(69.7\%\) of Seattle's.

Assume the cost is proportional to area. The original patio has area \(15 \times 20 = 300\text{ ft}^2\), and the new patio has area \(22 \times 25 = 550\text{ ft}^2\).

So the new patio is \(550 \div 300 = \tfrac{11}{6}\) times as large as the original patio.

Multiply the original cost by that factor: \(\$2{,}275 \times \tfrac{11}{6} \approx \$4170.83\). Rounded to the nearest dollar, the estimate is \(\$4{,}171\).

The population is the full group the study wants information about, and the sample is the part that was actually surveyed.

A parameter describes the population, while a statistic describes the sample.

So here the population is all district households, the sample is the 350 surveyed households, the parameter is the true percent of all district households that support the later start time, and the statistic is the sample result of 61%.

A simple random sample gives every individual an equal chance of selection. A stratified sample splits the population into groups first and samples from each group. A cluster sample randomly selects whole groups and surveys everyone in those groups. A voluntary-response sample relies on people choosing to participate.

That makes the answers: a) simple random, b) stratified, c) cluster, d) voluntary response.

A study can only support a conclusion about the larger population if the sample is reasonably representative of that population.

In part a, coupon users are not necessarily representative of all customers. In part b, Walmart shoppers are not necessarily representative of all Springfield residents. In part c, students who choose to reply to an online survey create voluntary-response bias.

So none of the three conclusions is justified as stated.

An observational study records what is already happening, while an experiment imposes a treatment.

Cause-and-effect conclusions require a well-designed experiment, not just observed associations.

So a) observational, no cause-and-effect conclusion; b) experiment, a cause-and-effect conclusion may be justified if the design is sound; c) observational, no cause-and-effect conclusion.

The treatment group is the group that receives the actual condition being tested, and the control group provides the comparison. A placebo looks like the treatment but has no active ingredient.

Since neither the volunteers nor the timer knows who got which drink, the study is double-blind.

Random assignment matters because it helps create comparable groups and reduces bias from pre-existing differences.

A confounding variable is another factor that may help explain the observed result, so the claimed cause is not established.

In part a, stronger study habits or general organization could explain both carrying a water bottle and higher scores. In part b, stress, workload, or reverse causation could explain the association between coffee and sleep. In part c, other changes such as policing, weather, or seasonal trends could affect crime.

So none of the three cause-and-effect claims is justified from the information given.

Count how many ages fall in each class.

There are 2 values in the 20s, 5 in the 30s, 2 in the 40s, 7 in the 50s, 6 in the 60s, 7 in the 70s, and 1 in the 80s.

Those frequencies add to 30, so the distribution is complete.

Add the bar heights: 5 + 4 + 4 + 3 + 1 + 2 = 19 adults.

The bar for 0 children has height 5, so the percentage is \(\frac{5}{19} \times 100 \approx 26.3\%\).

Fun is 17% of the total spending.

Compute 17% of $2800: \(0.17 \times 2800 = 476\). So Luciana spent $476 on Fun.

The class boundaries increase by 3 each time, so the class width is 3.

Add the frequencies to get the sample size: \(10 + 7 + 6 + 3 + 2 + 5 = 33\).

The mean gives the average score, so compare 8.4 and 7.5.

Class A scored better on average because 8.4 is larger. The smaller standard deviation, 0.3, means Class B's scores were more tightly clustered, so Class B was more consistent.

The mean is \(\frac{4+6+7+9+11+13+15+16+19}{9} = \frac{100}{9} \approx 11.1\).

With 9 values, the median is the middle value, so the median is 11. The lower half is 4, 6, 7, 9, so \(Q_1 = 6.5\). The upper half is 13, 15, 16, 19, so \(Q_3 = 15.5\).

That gives the five-number summary \((4, 6.5, 11, 15.5, 19)\). Using the sample standard deviation formula, the standard deviation is about \(5.0\).

The mean is \(\frac{3+5+6+8+9+11+12+14+15+17}{10} = \frac{100}{10} = 10\).

With 10 values, the median is the average of the two middle values: \(\frac{9+11}{2} = 10\). The lower half is 3, 5, 6, 8, 9, so \(Q_1 = 6\). The upper half is 11, 12, 14, 15, 17, so \(Q_3 = 14\).

That gives the five-number summary \((3, 6, 10, 14, 17)\). Using the sample standard deviation formula, the standard deviation is about \(4.6\).

The total number of responses is \(361 + 170 = 531\).

The probability of choosing a person who answered yes is \(\frac{361}{531}\), which is about \(0.680\) to three decimal places.

There are \(20\) marbles total.

For part (a), \(12\) of the \(20\) marbles are red, so \(P(\text{red}) = \frac{12}{20} = \frac{3}{5}\).

For part (b), the odd-numbered marbles are 1, 3, 5, and 7, 9, 11 among the red marbles and 1, 3, 5, 7 among the blue marbles, so \(P(\text{odd}) = \frac{10}{20} = \frac{1}{2}\).

For part (c), there are \(12\) red marbles, \(10\) odd-numbered marbles, and \(6\) marbles counted in both groups, so \(P(\text{red or odd}) = \frac{12 + 10 - 6}{20} = \frac{16}{20} = \frac{4}{5}\).

For part (d), the blue even-numbered marbles are 2, 4, 6, and 8, so \(P(\text{blue and even}) = \frac{4}{20} = \frac{1}{5}\).

Because two marbles are pulled at the same time, this is a without-replacement probability.

The probability the first marble is red is \(\frac{20}{54}\), and then the probability the second marble is red is \(\frac{19}{53}\). Multiply: \(\frac{20}{54} \cdot \frac{19}{53} = \frac{190}{1431} \approx 0.133\).

Let \(A\) be the event that the pedestrian was intoxicated and \(B\) be the event that the driver was not intoxicated.

Using the addition rule, \(P(A \text{ or } B) = \frac{298 + 840 - 235}{988} = \frac{903}{988} \approx 0.914\). As a percent, that is about \(91.4\%\).

The word given means you restrict the sample space first.

Among the male students, \(16\) earned a C out of \(53\) male students total. So \(P(C \mid \text{male}) = \frac{16}{53} \approx 0.302\).

Use expected value: multiply each outcome by its probability and add.

Here that gives \(4\left(\frac{1}{41}\right) + 3\left(\frac{10}{41}\right) + (-1)\left(\frac{30}{41}\right) = \frac{4}{41} \approx 0.10\). The expected value is a gain of about \(\$0.10\) per play.