The weight of oranges growing in an orchard is normally distributed with a mean weight of 5.5 oz. and a standard
deviation of 1 oz. What is the probability that a randomly
selected orange from the orchard weighs more than 4 oz., to the
nearest thousandth?

The probability is 0.933

Step-by-step explanation:

We start by calculating the z-score

Mathematically;

z-score = (x-mean)/SD

x = 4

mean = 5.5

SD = 1

z-score = (4-5.5)/1 = -1.5

So we proceed to get the probability

This is;

P( z > -1.5)

we can get this from the standard normal distribution table

That will be

P(z > -1.5) = 0.93319

To the nearest thousandth, this is 0.933

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.

step-by-step explanation:

their are no angles. sorry cant .