Answer:
18.75
Step-by-step explanation:
might not be 100% right
Answer:
Proportion of women having blood pressures between 88.1 and 89.4 is 3.99% or close to 4%.
Step-by-step explanation:
We are given that a recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6.
Let X = blood pressures of adult women in the United States
So, X ~ N(
)
The z score probability distribution is given by;
Z =
~ N(0,1)
where,
= population mean
= population standard deviation
So, Probability that women have blood pressures between 88.1 and 89.4 is given by = P(88.1 < X < 89.4) = P(X < 89.4) - P(X
88.1)
P(X < 89.4) = P(
<
) = P(Z < 1.05) = 0.85314
P(X
88.1) = P(
) = P(Z
0.89) = 0.81327
Therefore, P(88.1 < X < 89.4) = 0.85314 - 0.81327 = 0.0399 or approx 4%
Hence, proportion of women have blood pressures between 88.1 and 89.4 is 4%.
If you expand -9(-2x-3) you get 18x+27 (a minus times a minus is a positive) so the answer is A.
Answer:
4
Step-by-step explanation:
4 num nuts