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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%.
5x + 3 
23
5x 20
x 4
So it can be {4, 8, 12}
If this helps, can you give me the brainliest answer? I need it...
Thank you very much.
5x
x
So it can be {4, 8, 12}
If this helps, can you give me the brainliest answer? I need it...
Thank you very much.