John drives to work each morning, and the trip takes an average of µ = 38 minutes. the distribution of driving times is approximately normal with a standard deviation of σ = 5 minutes. for a randomly selected morning, what is the probability that john’s drive to work will take less than 35 minutes?

0.2742531

Step-by-step explanation:

Let x denote the random variable that represents the driving times.

As per given we have,

Z-score value corresponds to x= 35,

Using z-value table ,

The probability that John’s drive to work will take less than 35 minutes:-

P(X<35)=P(z<-0.6)=1-P(z<0.6)=1-0.7257469=0.2742531

hence, the required probability = 0.2742531

the answer is. -82562234434584

step-by-step explanation:

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !