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!!!
Okay so first we need to find how many days are in March and February. March has 31 days and because this year was a leap year February has 29 days.
The next step is to convert days to hours.
March: 31x24=744
February: 29x24=696
Now its time to graph
Answer:
Jackie sold 12 cars.
Step-by-step explanation:
If we call the number of cars Oscar sold O, and the number of cars Jackie sold J, we can say the following:
O = J + 6
As Oscar sold 6 cars more than Jackie.
Together, they sold 30 cars.
O + J = 30
Since we know that:
O = J + 6
... we can put this into our previous equation.
O + J = 30
(J + 6) + J = 30
J + J + 6 = 30
2 * J + 6 = 30
Subtract 6 from both sides:
2 * J = 24
Divide both sides by 2:
J = 24 / 2
J = 12
Jackie sold 12 cars.
I'm assuming that the full statement is ΔABC is congruent to ΔDEF.
Given:
m∠A = 70
m∠B = 47
m∠C = 63
Since both triangles are congruent, the corresponding angles in ΔDEF also has the same measure as those in ΔABC.
m∠D = 70
m∠E = 47
m∠F = 63
The measure of angle F is 63.
we are given

Firstly, we will split terms

We can group first two and last two terms

we can see that 2x^2 is common in first two terms
and 3 is common in last two terms
so, we can factor factor out 2x^2 from first group and 3 from second group
we get

we can see that each terms are multiple of 6x-1
so, we can factor out 6x-1
so, we get
............Answer