a missile is fired with 96% accuracy. Each firing is independent of the other. If four missiles are fired what is the probability only the second one misses

0.03538944

Step-by-step explanation:

Given, The accuracy of missile = 96%

i.e. The probability that it will hit the shot = 0.96

Probability of missing the shot = 1-0.96= 0.04

Now, each missile is independent of other.

Therefore, the probability only the second one misses = (probability that first will hit the shot) x (probability that second one misses ) x (probability that third will hit the shot)x (probability that fourth will hit the shot)

Thus, the probability only the second one misses is

Hence, the required probability = 0.03538944

94

step-by-step explanation:

average the numbers and use a variable to figure out the minimum to get a 90.

(79+94+91+92+x)/5=90

x = 94

$3,200 x 7.25% = $3,200 x .0725 = $232 interest monthly

$232 monthly x 12 annual = $2, 784

step-by-step explanation: