Answer:
all work is shown and pictured
Answer:
Step-by-step explanation
Hello!
Be X: SAT scores of students attending college.
The population mean is μ= 1150 and the standard deviation σ= 150
The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.
If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:
H₀: μ = 1150
H₁: μ ≠ 1150
α: 0.05
![Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~~N%280%3B1%29)

The p-value for this test is 0.0949
Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.
I hope it helps!
Mark points (0,0), (-1,-1), (1,-1), (-2,-4), (2,-4)
Answer:
It would be 5
Step-by-step explanation:
29. A carpenter worked for 4 days to finish a repair job. She charged $825 for 15 hours plus $278 for materials. What is the carpenter's rate per hour?
Let’s start solving:
=> 825 dollars + 278 dollars = 1 103 dollars is the total payment a carpenter have received
=> 15 hours is the total hours he worked
=> 1 103 / 15 = 73.5 dollars per hour. This is already including the material or equipment fee.