Estimate the percent using decimals<br> 52% of 71

Geometry and perimeter help plz :(

Novay_Z [31]

Answer:

11.)4.79mi

12.)8cm

13.)10.6km

14.)5.93

15.)2yd

16.)5.41in

17)2y

18)2m

19)11.8yd

20)4.51km

Step-by-step explanation:

11.)15.8x2=31.6 31.6/6.6=4.79

12.)64/8=8

13.)26x2=54 54/4.9=10.6

14.)8.6x2=17.2 17.2/5.93

15.)2/1=2

16.)29.2/5.4=5.41

17)4/2=2

18)10/5=2

19)139.2/11.8=11.8

20)20.3/4.5=4.51

3 0

3 years ago

A running coach wants to know if participating in weekly running clubs significantly improves the time to run a mile. The runnin

patriot [66]

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!!!

7 0

4 years ago

Jordi trains on his bicycle by riding for a total of 300 miles each week

Katen [24]

Y = 300 (miles) * x (independent variable)
I hope this satisfies you.

7 0

4 years ago

Suppose you were to draw all possible samples of size 36 from a large population with a mean of 650 and a standard deviation of

Lady_Fox [76]

Answer:

By the Central Limit Theorem, it is approximately normal with mean 650 and standard deviation 4.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .