The student scores on a history test are normally distributed with an unknown population mean and standard deviation A random sample of 38 scores is taken and results in a sample mean of 85 points and sample standard deviation of 4 points. dr 0,10 0.0 0.025 0.01 0.005 3 1.306 1.690 2.030 2.438 2.724 1.306 1.688 2.028 2.434 2.719 1.305 1687 2.026 2.431 2.715 1.304 1.686 2.024 2.429 2.712 1.304 1685 2.023 2.426 2.708 Find a 90% confidence interval estimate for the population mean using the Student's distribution . Round the final answers to two decimal places.

The 90% confidence interval = (83.9, 86.1)

Step-by-step explanation:

The formula for confidence interval =

Mean ± z × standard deviation/√n

From the question:

n = 38

Mean = 85 points

Standard deviation = 4 points

Z score for 90% confidence interval = 1.645

85 ± 1.645 × 4/√38

85 ± 1.067416951

Confidence Interval =

85 - 1.067416951

= 83.932583049

≈ 83.9

85 + 1.067416951

= 86.067416951

≈ 86.1

Therefore, the 90% confidence interval = (83.9, 86.1)

sorry, i don't know i just need points, i'm new here. chao

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: