The scores on a certain test are normally distributed with a mean score of 69 and a standard deviation of 4. what is the probability that a sample of 90 students will have a mean score of at least 69.4216? your answer should be a decimal rounded to the fourth decimal place.

Solution: We are given:

μ=69,σ=4,n=90

We have to find P(Mean≥69.4216)

We need to first find the z score

z= (xbar-μ)/(σ/sqrt(n))

=(69.4216-69)/(4/sqrt(90))

=1.00

Now we have to find P(z≥1.00)

Using the standard normal table, we have:

P(z≥1.00)= 0.1587

Therefore the probability that a sample of 90 students will have a mean score of at least 69.4216 is 0.1587.

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3

the answer is b.

surface area: 169π

volume: 366π

i got it down.