Answer:
Option B:


Classification:
The hypothesis test is Two-tailed.
Step-by-step explanation:
The mean length of imprisonment for motor-vehicle theft offenders in this country is 22.1 months.
This means that the null hypothesis is that the mean is of 22.1 months, that is:

A hypothesis test is to be performed to determine whether the mean length of imprisonment for motor-vehicle theft offenders in this city differs from the national mean of 22.1 months.
At the alternate hypothesis, we test if this mean is different of 22.1, that is:

Which means that the answer is given by option b).
Which of the following is the correct classification of the hypothesis test?
We test if the mean is different from a value, which means that the hypothesis test is Two-tailed.
Answer is A
X-3/x^2-4 has two excluded values 2, -2
5.6 because 7=1=8 so 8-2.4=5.6
hope this helped please leave a thank you
Answer:
A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.
B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.
C. The expected value of X is 6.75, and the standard deviation of X is 2.17.
Step-by-step explanation:
The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.
With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

A. P(x=6)

B. P(x≥10)




C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:

<h3>
Answer: 1/2 (choice A)</h3>
===========================================
Explanation:
The two equations given to us are
Divide the second equation over the first equation and that would lead to b = 6
Notice how the 'a' terms divide to 1 and go away, i.e. cancel out.
The b terms divide to (b^2)/b = b
The right hand side values divide to 18/3 = 6
So that's how we end up with b = 6
-------------------------
Now if b = 6, then we can say,
ab = 3
a*6 = 3
a = 3/6
a = 1/2
Or we could say
ab^2 = 18
a*6^2 = 18
a*36 = 18
a = 18/36
a = 1/2