A new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. What is the probability that you test exactly 4 batteries

The probability that you test exactly 4 batteries is 0.0243.

Step-by-step explanation:

We are given that a new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found.

Suppose that 90% of all batteries have acceptable voltages.

Let the probability that batteries have acceptable voltages = P(A) = 0.90

So, the probability that batteries have unacceptable voltages = P(U) = 1 - P(A) = 1 - 0.90 = 0.10

Now, the probability that you test exactly 4 batteries is given by the three cases. Firstly, note that batteries will be tested until two acceptable ones have been found.

So, the cases are = P(AUUA) + P(UAUA) + P(UUAA)

This means that we have tested 4 batteries until we get two acceptable batteries.

So, required probability = (0.90 0.10 0.10 0.90) + (0.10 0.90 0.10 0.90) + (0.10 0.10 0.90 0.90)

     =  0.0081 + 0.0081 + 0.0081 = 0.0243

Hence, the probability that you test exactly 4 batteries is 0.0243.

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50