Costs for standard veterinary services at a local animal hospital follow a normal distribution with a mean of $88 and a standard deviation of $24. what is the probability that one bill for veterinary services costs less than $42? a) 0.9862b) 0.9724c) 0.5000d) 0.0138e) 0.0276f) none of the above

the correct option is f

Step-by-step explanation:

Since costs for standard veterinary services at a local animal hospital follow a Normal distribution, we would apply the the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = costs for standard veterinary services at a local animal hospital

u = mean cost

s = standard deviation

From the information given,

u = $88

s = $24

We want to determine the probability that one bill for veterinary services costs less than $42. It is expressed as

P(x lesser than or equal to 42)

For x = 42,

z = (42 - 88)/24 =- 1.92

Looking at the normal distribution table, the corresponding value for the z score is 0.02743

P(x lesser than or equal to 42) = 0.02743

160

step-by-step explanation:

idk

i hope i

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right