The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 267 and a standard deviation of 15. What percentage of pregnancies last beyond 246 days? P(X > 246 days) =

91.92% of pregnancies last beyond 246 days

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

What percentage of pregnancies last beyond 246 days?

We have to find 1 subtracted by the pvalue of Z when X = 246. So

has a pvalue of 0.0808

1 - 0.0808 = 0.9192

91.92% of pregnancies last beyond 246 days

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

answer: the resulting cross section is a rectangle. solution: a plane parallel to the base of a triangular prism will intersect a cross section that is the same shape as its bases. so the cross section is a triangle.

step-by-step explanation: