The correct answer is 6. The volume of a cylinder is πr²h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Since the storage silo is a cylinder with volume 72π cubic yards and height 8 yards, it is true that 72π = πr² (8), where r is the radius of the base of the cylinder, in yards. Dividing both sides of 72π = πr² (8) by 8π gives r² = 9, and so the radius of base of the cylinder is 3 yards. Therefore, the diameter of the base of the cylinder is 6 yards.

The correct answer is 96. Since each day has a total of 24 hours of time slots available for the station to sell, there is a total of 48 hours of time slots available to sell on Tuesday and Wednesday. Each time slot is a 30-minute interval, which is equal to a ¹⁄₂ -hour interval. Therefore, there are a total of \(\frac{48\, hours}{\frac{1}{2}hours/time\, slot}\)= 96 time slots of 30 minutes for the station to sell on Tuesday and Wednesday.

The correct answer is ⁵⁄₈ or .625. Based on the line graph, the number of portable media players sold in 2008 was 100 million, and the number of portable media players sold in 2011 was 160 million. Therefore, the number of portable media players sold in 2008 is \(\frac{100\, million}{160\, million}\) of the portable media players sold in 2011. This fraction reduces to ⁵⁄₈. Either ⁵⁄₈ or its decimal equivalent, .625, may be gridded as the correct answer.

The correct answer is 107. Since the weight of the empty truck and its driver is 4500 pounds and each box weighs 14 pounds, the weight, in pounds, of the delivery truck, its driver, and x boxes is 4500 + 14x. This weight is below the bridge’s posted weight limit of 6000 pounds if 4500 + 14x < 6000. That inequality is equivalent to 14x ≤ 1500 or x < [latex]\frac{1500}{14}[/latex] = 107 ¹⁄₇ . Since the number of packages must be an integer, the maximum possible value for x that will keep the combined weight of the truck, its driver, and the x identical boxes below the bridge’s posted weight limit is 107.

The correct answer is any number between 4 and 6, inclusive. Since Wyatt can husk at least 12 dozen ears of corn per hour, it will take him no more than \(\frac{72}{12}\)=6 hours to husk 72 dozen ears of corn. On the other hand, since Wyatt can husk at most 18 dozen ears of corn per hour, it will take him at least \(\frac{72}{18}\)= 4 hours to husk 72 dozen ears of corn. Therefore, the possible times it could take Wyatt to husk 72 dozen ears of corn are 4 hours to 6 hours, inclusive. Any number between 4 and 6, inclusive, can be gridded as the correct answer.

Any quadratic function q can be written in the form q(x) = a(x − h)² + k, where a, h, and k are constants and (h, k) is the vertex of the parabola when q is graphed in the coordinate plane. (Depending on the sign of a, the constant k must be the minimum or maximum value of q, and h is the value of x for which a(x − h)² = 0 and q(x) has value k.) This form can be reached by completing the square in the expression that defines q. The given equation is y = x² − 2x − 15, and since the coefficient of x is −2, the equation can be written in terms of (x − 1)² = x² − 2x + 1 as follows: y = x² − 2x − 15 = (x² − 2x + 1) − 16 = (x − 1)2 − 16. From this form of the equation, the coefficients of the vertex can be read as (1, −16)

Choices A and C are incorrect because the coordinates of the vertex A do not appear as constants in these equations. Choice B is incorrect because it is not equivalent to the given equation.

If the polynomial p(x) is divided by x − 3, the result can be written as \(\frac{p(x)}{x − 3}\) = q(x) + \(\frac{r}{x − 3}\), where q(x) is a polynomial and r is the remainder. Since x − 3 is a degree 1 polynomial, the remainder is a real number. Hence, p(x) can be written as p(x) = (x − 3)q(x) + r, where r is a real number. It is given that p(3) = −2 so it must be true that −2 = p(3) = (3 − 3)q(3) + r = (0)q(3) + r = r. Therefore, the remainder when p(x) is divided by x − 3 is −2.

Choice A is incorrect because p(3) = −2 does not imply that p(5) = 0. Choices B and C are incorrect because the remainder −2 or its negative, 2, need not be a root of p(x).

To determine which quadrant does not contain any solutions to the system of inequalities, graph the inequalities. Graph the inequality y ≥ 2x + 1 by drawing a line through the y-intercept (0, 1) and the point (1, 3), and graph the inequality y > ¹⁄₂x − 1 by drawing a dashed line through the y-intercept (0, −1) and the point (2, 0), as shown in the figure below.

The solution to the system of inequalities is the intersection of the shaded regions above the graphs of both lines. It can be seen that the solutions only include points in quadrants I, II, and III and do not include any points in quadrant IV.

Choices A and B are incorrect because quadrants II and III contain solutions to the system of inequalities, as shown in the figure above. Choice D is incorrect because there are no solutions in quadrant IV.

The area of the field is 100 square meters. Each 1-meterby-1-meter square has an area of 1 square meter. Thus, on average, the earthworm counts to a depth of 5 centimeters for each of the regions investigated by the students should be about \(\frac{1}{100}\) of the total number of earthworms to a depth of 5 centimeters in the entire field. Since the counts for the smaller regions are from 107 to 176, the estimate for the entire field should be between 10,700 and 17,600. Therefore, of the given choices, 15,000 is a reasonable estimate for the number of earthworms to a depth of 5 centimeters in the entire field.

Choice A is incorrect; 150 is the approximate number of earthworms in 1 square meter. Choice B is incorrect; it results from using 10 square meters as the area of the field. Choice D is incorrect; it results from using 1,000 square meters as the area of the field.

Let x represent the number of pears produced by the Type B trees. Then the Type A trees produce 20 percent more pears than x, which is x + 0.20x = 1.20x pears. Since Type A trees produce 144 pears, the equation 1.20x = 144 holds. Thus x =\(\frac{144}{1.20}\)= 120. Therefore, the Type B trees produced 120 pears.

Choice A is incorrect because while 144 is reduced by approximately 20 percent, increasing 115 by 20 percent gives 138, not 144. Choice C is incorrect; it results from subtracting 20 from the number of pears produced by the Type A trees. Choice D is incorrect; it results from adding 20 percent of the number of pears produced by Type A trees to the number of pears produced by Type A trees.