Asample of 4 different calculators is randomly selected from a group containing 17 that are defective and 37that have no defects. what is the probability that at least one of the calculators is defective?

0.8025

Step-by-step explanation:

Given : The number of defective calculators : 17

The number of calculators are not defective : 37

Total calculators : 37+17=54

The probability of the calculators are defective :

Binomial distribution formula :-

, where P(x) is the probability of success in x trials, n is total trials and p is the probability of success for one trial.

The probability that at least one of the calculators is defective is given by :-

d

step-by-step explanation:

by observation, we can see that the the graph is simply the graph of y = |x| that has been shifted 4 units downwards and 3 units to the left.

we can recognize in choice d :

y = | x + 3 |   - 4 (rewriting in general form)

y = | x - (- 3) |   - 4

the "-4 " term means   that the graph has been shifted by 4 units in the negative y direction

the "-3 " term means   that the graph has been shifted by 3 units in the negative x direction

this matches our observation of the graph described above.

answer: x=60