Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,750 pounds and the standard deviation is 260 pounds. Assume that the population follows the normal distribution. Forty-five trucks are randomly selected and weighed. Within what limits will 99% of the sample means occur?

99% of the sample means will occur between 5,650.20 pounds and 5,849.80.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 5,750 - 99.80 = 5,650.20

The upper end of the interval is the mean added to M. So it is 99.80 + 5,750 = 5,849.80

99% of the sample means will occur between 5,650.20 pounds and 5,849.80.

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

so the possibilities of the measurement of the side of the triangle are either it is longest side or not the longest side

when it is the longest side

s < 25 + 35s < 60

if it is not the longest side

35 < 25 + s10 < sso 10 < s < 60so the perimeter70 < p < 130