Answer:
Option B is correct.
Use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.
Step-by-step Explanation:
The clear, complete table For this question is presented in the attached image to this solution.
It should be noted that For this question, the running coach wants to test if participating in weekly running clubs significantly improves the time to run a mile.
In the data setup, the mean time to run a mile in January for those that participate in weekly running clubs and those that do not was provided.
The mean time to run a mile in June too is provided for those that participate in weekly running clubs and those that do not.
Then the difference in the mean time to run a mile in January and June for the two classes (those that participate in weekly running clubs and those that do not) is also provided.
Since, the aim of the running coach is to test if participating in weekly running clubs significantly improves the time to run a mile, so, it is logical that it is the improvements in running times for the two groups that should be compared.
Hence, we should use the difference in sample means (10 and 8) in a hypothesis test for a difference in two population means.
Hope this Helps!!!
Let us say that:
o = cost of oranges per pound
p = cost of pears per pound
so that:
o = p – 2
Therefore:
10o + 8p = 61
10 (p – 2) + 8p = 61
10p – 20 + 8p = 61
18p = 81
p = 4.5
p = $4.5 per pound
So 3 pounds of pears would cost:
total cost = 3 * 4.5
total cost = $13.5
I believe that this problem has the following choices:
It must be equal to BQ .<span>
It must be wider than when he constructed the arc centered at
point A.
It must be equal to AB .
It must be the same as when he constructed the arc centered
at point A.</span>
The correct answer is the last one:
It must be the same as when he constructed the arc centered
at point A.
<span> </span>
Answer:
Step-by-step explanation:
150/x=sin 15
x=150/sin 15 ≈579.56 m
Answer:
4 nickels
12 dimes
Step-by-step explanation:
Dimes are worth .1 each while nickels are .05 each.
We have 8 more dimes than nickels. Let d represent number of dimes and n for number of nickels. This means we have d=8+n.
If all our nickels and dimes together are worth 1.4 then we have another equation .1d+.05n=1.4
Lets put our equations together:
d=8+n
.1d+.05n=1.4
‐-----------------Plug first equation into second.
.1(8+n)+.05n=1.4
Distribute
.8+.1n+.05n=1.4
Combine like terms
.8+.15n=1.4
Subtract .8 on both sides
.15n=.6
Divide both sides by .15
n=.6/.15=4
Remember there are 8 more dimes so d=8+4=12.