Answer:
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the 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 = 25 - 1 = 24
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.064
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4.21 = 77.29
The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Hey there!
The x-intercept is where the line crosses the x-axis. The x-axis is horizontal. It crosses the line three left from the origin, or (0,0). This is -3. The x-intercept is -3. The answer would be either B or C. The line never hits the y-axis, so the y-intercept is none. The answer is C.
I hope this helps!
X=-1.8
To solve first use the distributive property to change the equation to -2x+1 1/3+1=5
Then you subtract 1 from 5 so then the equation looks like this: -2x+1 1/3=4
Next subtract 1 1/3 from 4 to get 3.6
then divide it by -2 and you get -1.8
Measure the lengths of the runways in inches and then times your answer by 700 and then round that answer.
2y-x=-8.
Minus= -
equals is =.