The weight of oranges growing in an orchard is normally distributed with a mean weight of 5 oz. and a standard deviation of 0.5 oz. From a batch of 1500 oranges, how many would be expected to weight less than 5 oz., to the nearest whole number?

The interval that would represent weights of the middle 95% of all oranges from this orchard is from 5 oz to 7 oz.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 6

Standard deviation = 0.5

Middle 95% of weights:

By the Empirical Rule, within 2 standard deviations of the mean.

6 - 2*0.5 = 5

6 + 2*0.5 = 7

The interval that would represent weights of the middle 95% of all oranges from this orchard is from 5 oz to 7 oz.

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

answer:

your answer is: c. 196 7/8 in^3.