Test (With calculator) - 3 - SAT MCQ

The probability that a person from Group Y who recalled at least 1 dream was chosen at random from the group of all people who recalled at least 1 dream is equal to the number of people in Group Y who recalled at least 1 dream divided by the total number of people in the two groups who recalled at least 1 dream. The number of people in Group Y who recalled at least 1 dream is the sum of the 11 people in Group Y who recalled 1 to 4 dreams and the 68 people in Group Y who recalled 5 or more dreams: 11 + 68 = 79. The total number of people who recalled at least 1 dream is the sum of the 79 people in Group Y who recalled at least 1 dream, the 28 people in Group X who recalled 1 to 4 dreams, and the 57 people in Group X who recalled 5 or
more dreams: 79 + 28 + 57 = 164. Therefore, the probability is 79/164.
Choice A is incorrect; it is the probability of choosing at random a person from Group Y who recalled 5 or more dreams. Choice B is incorrect; it is the probability of choosing at random a person from Group Y who recalled at least 1 dream. Choice D is incorrect; it is the probability of choosing at random a person from the two groups combined who recalled at least 1 dream.

Choice B is correct. The amounts given in the table are in thousands of dollars. Therefore, the amount in the annual budget for agriculture/natural resources is actually $488,106,000 in 2010 and $358,708,000 in 2008. Therefore, the change in the budgeted amount is $488,106,000 - $358,708,000 = $129,398,000. Hence, the average change in the annual budget for agriculture/natural resources from 2008 to 2010 is . Of the options given, this average rate of change is closest to $65,000,000 per year.
Choices A and C are incorrect and may result from errors in setting up or calculating the average rate of change. Choice D is incorrect; $130,000,000 is the approximate total change in the annual budget for agriculture/natural resources from 2008 to 2010, not the average rate of change from 2008 to 2010.

The human resources budget in 2007 was 4,051,050 thousand dollars, and the human resources budget in 2010 was 5,921,379 thousand dollars. Therefore, the ratio of the 2007 budget to the 2010 budget is slightly greater than 4/6 = 2/3. Similar estimates
for agriculture/natural resources give a ratio of the 2007 budget to the 2010 budget of slightly greater than 4; for education, a ratio of slightly greater than 3/4; for highways and transportation, a ratio of slightly less than 2/3; and for public safety, a ratio of slightly greater than 5/9.
Therefore, of the given choices, education’s ratio of the 2007 budget to the 2010 budget is closest to that of human resources.
Choices A, C, and D are incorrect because the ratio of the 2007 budget to 2010 budget for each of the programs given in these choices is further from the corresponding ratio for human resources than the corresponding ratio for education.

Choice A is correct. The equation of a circle can be written as (x - h)² + (y - k)² = r² where (h, k) are the coordinates of the center of the circle and r is the radius of the circle. Since the coordinates of the center of the circle are (0, 4), the equation of the circle is x² + (y - 4)² = r². The radius of the circle is the distance from the center,(0, 4), to the given endpoint of a radius, ( 4/3, 5). By the distance formula,  Therefore, an equation of the given circle is 
Choices B and D are incorrect. The equations given in these choices represent a circle with center (0, -4), not (0, 4). Choice C is incorrect; it results from using r instead of r² in the equation for the circle.

When the ball hits the ground, its height is 0 meters. Substituting O for h in h = -4.9t² + 25t gives 0 = -4.9t² + 25t, which can be rewritten as O = t (-4.91 + 25). Thus, the possible values of t are t = 0 and t =  The time t = 0 seconds corresponds to the time the ball is launched from the ground, and the time t ≈ 5.1 seconds corresponds to the time after launch that the ball hits the ground. Of the given choices, 5.0 seconds is closest to 5.1 seconds, so the ball returns to the ground approximately 5.0 seconds after it is launched.
Choice A, B, and C are incorrect and could arise from conceptual or computation errors while solving 0 = -4.91² + 25t for t.
Choices B and D are incorrect. The equations given in these choices represent a circle with center (0, -4), not (0, 4). Choice C is incorrect; it results from using r instead of r² in the equation for the circle.

Choice B is correct. Let x represent the number of pears produced by the Type B trees. Type A trees produce 20 percent more pears than Type B trees, or x, which can be represented as x + 0.20x = 1.20x pears.
Since Type A trees produce 144 pears, it follows that 1.20x = 144. Thus x = 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.

Choice C is correct. The area of the held is 100 square meters. Each 1-meter-by-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 1/100 of the total number of earthworms to a depth of 5 centimeters in the entire held. Since the counts for the smaller regions are from 107 to 176, the estimate for the entire held 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 held.
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 held. Choice D is incorrect; it results from using 1,000 square meters as the area of the held.

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), as shown. The solutions to this inequality are all points contained on and above this line. Graph the inequality  drawing a dashed line through the y-intercept (0, -1) and the point (2, 0), as shown. The solutions to this inequality are all points above this dashed line.

The solution to the system of inequalities is the intersection of the 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.

Choice D is correct. If the polynomial p(x) is divided by x - 3, the result can be written as 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 opposite, 2, need not be a root of p(x).

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. This form can be reached by completing the square in the expression that defines q. The equation of the graph is y = x² - 2x - 15.
Since the coefficient of x is -2, this 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.

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 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 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.

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. Subtracting 4500 from both sides of this inequality and then dividing both sides by 14 yields 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.

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 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, maybe gridded as the correct answer.

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 = 96 time slots of 30 minutes for the station to sell on Tuesday and Wednesday.

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 follows that 72π = πr²(8), where r is the radius of the base of the cylinder, in yards. Dividing both sides of the equation 72π = πr²(8) by 8π gives r² = 9, and so the radius of the base of the cylinder is 3 yards. Therefore, the diameter of the base of the cylinder is 6 yards.