An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiveness of the new method was evaluated in a simulation study. In 50 simulations, the mean improvement in traffic flow in a particular intersection was 654.7 vehicles per hour, with a standard deviation of 311.7 vehicles per hour. Find a 95% confidence interval for the improvement in traffic flow due to the new system. Round the answers to three decimal places. The 95% confidence interval is ( , ).

The 95% confidence interval is (28.183, 1281.217).

Step-by-step explanation:

We have the sample's standard deviation, so we use the students' t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 50 - 1 = 49

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.01

The margin of error is:

M = T*s = 2.01*311.7 = 626.517

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 654.7 - 626.517 = 28.183 vehicles per hour

The upper end of the interval is the sample mean added to M. So it is 654.7 + 626.517 = 1281.217 vehicles per hour

The 95% confidence interval is (28.183, 1281.217).

the answer is a

step-by-step explanation:

39.00 x 24 = total installment price

carrying installment price - cash price

needed to save the .00/

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet