Afield test for a new exam was given to randomly selected seniors. the exams were graded, and the sample mean and sample standard deviation were calculated. based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 5% of 75%. is the confidence interval at 90%, 95%, or 99%? what is the margin of error? calculate the confidence interval and explain what it means in terms of the situation.

90% is the answer for confidence interval

Step-by-step explanation:

Given that a field test for a new exam was given.75±0.05(75)

i.e. margin of error =0.05(75)=

The population under study was the seniors. Sample mean and std dev were calculated

Stated that the exam creator ased on the results, that on the same exam, 9 out of 10 seniors will have an average score within 5% of 75%

This implies that 9 out of 10 means in percent 90%

Hence the confidence interval is 90%

SInce 5% of 75 is given the range is 75±(75)

Margin of error =3.75

Confidence interval =(75±3.75)

=(71.25,78.75)

We can interpret this we are 90% confident that  as sample sizes increase, and samples are drawn at random the average grades of seniors would lie between 71.25 and 78.75

x=6 and x=3

beep beep imma sheep

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~