A company producing tennis racquets found that 3% of the racquets they produce are defective. The racquets are sent to shops in boxes of 6 and the shops send back boxes with more than one defective racquet. If a box is chosen at random and opened, what is the probability that it is sent back

0.82

Step-by-step explanation:

The question says that three percent (3 out of 100) of the rackets (racquets) produced by the company are found to be defective.

First thing to calculate is the probability that 1 in a box of 6 rackets are defective.

That probability is gotten by cross multiplying, as illustrated below;

3 - 100

x - 6

x = (6×3)/100 = 18/100 = 0.18

Since the probability of having/finding 1 defective racket in a box of 6 is 0.18, the probability of finding more than 1 (minimum of 2 and maximum of all six) defective rackets in same box would be (1 - 0.18) = 0.82

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

3

step-by-step explanation:

divide 13.5 by 9 to get the scale factor. it equals 1.5

2 * 1.5 = 3