The weight of oranges growing in an orchard is normally distributed with a mean
weight of 5 oz. and a standard deviation of 1 oz. Using the empirical rule, what
percentage of the oranges from the orchard weigh between 3 oz. and 7 oz.?

The percentage is 95.45%

Step-by-step explanation:

In this question, we are tasked with calculating the percentage of oranges weighing between a particular range.

To do this , we use the z-scores of both numbers in the range.

Mathematically, z-score can be calculated as (x - Mean)/SD

Thus we have P(3 <x < 7) = P( (3-5)/1 < z < (7-5)/1)

= P( -2 < z < 2)

Using standard score table, we have

P(-2<z<2) = 0.9545

The percentage is thus 9545/100 = 95.45

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1