Seventy-five boxes of light bulbs were examined during a quality control check. The number of defective light bulbs per box and the corresponding frequency are given in the table.Quantity AThe average (arithmetic mean) number of defective light bulbs per box for the 75 boxesQuantity BThe median number of defective light bulbs per box for the 75 boxes
Answer(s): B
To calculate the average number of defective bulbs, we first compute the total number of defective bulbs in all 75 boxes:For 0 defective bulbs: 35 × 0 = 0For 1 defective bulb: 20 × 1 = 20For 2 defective bulbs: 12 × 2 = 24For 3 defective bulbs: 2 × 3 = 6For 4 or more defective bulbs: 6 × 4 (using 4 as an estimate) = 24Now, add all these values together:0 + 20 + 24 + 6 + 24 = 74The total number of defective bulbs is 74, and the total number of boxes is 75. So, the average number of defective bulbs per box is:The median is the middle value in the data set when ordered. To find the median, we first determine the cumulative frequency:Cumulative frequency for 0 defective bulbs: 35Cumulative frequency for 1 defective bulb: 35 + 20 = 55Cumulative frequency for 2 defective bulbs: 55 + 12 = 67Cumulative frequency for 3 defective bulbs: 67 + 2 = 69Cumulative frequency for 4 or more defective bulbs: 69 + 6 = 75There are 75 boxes, so the median is the 38th value. From the cumulative frequency, we see that the 38th value falls in the range where there are 1 defective bulb, since the cumulative frequency for 0 defective bulbs is 35 and for 1 defective bulb is 55.Thus, the median number of defective bulbs is 1.Compare Quantity A and Quantity BQuantity A: The average number of defective bulbs per box is approximately 0.987.Quantity B: The median number of defective bulbs per box is 1.Since 0.987 is less than 1, Quantity B is greater.
Quantity AThe number of different prime factors of 500Quantity BThe number of different prime factors of 360
To compare the number of different prime factors of 500 and 360, we first need to find the prime factorization of each number.For 500, start by dividing by 2:500 ÷ 2 = 250250 ÷ 2 = 125Next, divide by 5:125 ÷ 5 = 2525 ÷ 5 = 55 ÷ 5 = 1So, the prime factorization of 500 is:500 = 22 × 53The distinct prime factors are 2 and 5, meaning there are 2 different prime factors.For 360, start by dividing by 2:360 ÷ 2 = 180180 ÷ 2 = 9090 ÷ 2 = 45Next, divide by 3:45 ÷ 3 = 1515 ÷ 3 = 5Finally, divide by 5:5 ÷ 5 = 1The prime factorization of 360 is:360 = 23 × 32 × 5.The distinct prime factors are 2, 3, and 5, meaning there are 3 different prime factors.Now, comparing the two quantities:The number of different prime factors of 500 is 2.The number of different prime factors of 360 is 3.Since 2 is less than 3, the number of different prime factors of 360 is greater.
Point P is shown in the rectangular coordinate system.Quantity AThe ratio of the x-coordinate of P to the y-coordinate of PQuantity B-1
Answer(s): C
Point P is in the second quadrant, and it is located along a dashed line that forms a 45-degree angle with the negative x-axis. In this case, the coordinates of P can be represented as (x, y).Since the angle between the line and the negative x-axis is 45 degrees, the coordinates of P must satisfy the property that the absolute values of x and y are equal. Given that the point is in the second quadrant, the x- coordinate is negative and the y-coordinate is positive. This means x = -a and y = a for some positive value a.Now, calculating the ratio of the x-coordinate to the y-coordinate: x divided by y = (-a) divided by a = -1.Thus, Quantity A is -1. Quantity B is also given as -1.
1 kilogram is approximately equal to 2.2 pounds.x < yQuantity AThe number of pounds equal to x kilogramsQuantity BThe number of kilograms equal to y pounds
Answer(s): D
The number of pounds equal to xxx kilograms can be found by multiplying x by 2.2:Quantity A = 2.2xTo convert y pounds to kilograms, we divide y by 2.2:Since we are only given x < y, it does not directly establish a clear comparison between quantities because of the conversion factor. The inequality x < y does not necessarily imply that the converted values will have a strict relationship.
A box contains 10 red balls, 5 blue balls, and no other balls. Ann will randomly select, without replacement, 2 balls from the box.Quantity AThe probability that Ann will select 2 red ballsQuantity B
The total number of balls in the box is:10(red balls) + 5(blue balls) = 15(total balls)The total number of ways to select 2 balls from 15 is given by the combination formula:Now, the number of ways to select 2 red balls from the 10 red balls is:Thus, the probability that Ann will select 2 red balls is:Since both quantities are equal, we conclude that the two quantities are the same.
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