Answer:
The first one is 36 i think and the second is 28 i think
Step-by-step explanation:
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!
121/100 is the answer. it cant be simplified anymore
Answer:
f(x)^-1 = x - 4
D: {x is all real numbers}
Step-by-step explanation:
change f(x) to y
switch x with y and vice versa
solve for y