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Suppose a normal distribution has a mean of 120 and a standard deviation of 10. What is P(x ≥ 130)?

A.
0.475

B.
0.84

C.
0.16

D.
0.975

B

Step-by-step explanation:

First thing to do here is to calculate the z-score

Mathematically;

z-score = (x-mean)/SD

here, x = 130, mean = 120 and SD = 10

Substituting these values

z-score = (130-120)/10 = 10/10 = 1

So the probability we want to calculate is;

P(z ≥ 1) = 1 - P(z <1)

From standard score table, P(z <1) = 0.15866

P(z ≥ 1) = 1-0.15866 = 0.84134

answer: b.48 for

step-by-step explanation:

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584