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7nadin3 [17]
3 years ago
5

Imagine that you are doing an exhaustive study on the children in all of the daycares in your school district. You are particula

rly interested in how much time children spend in active play on weekends. You find that for this population of 2,431 children, the average number of minutes spent in active play on weekends is μ = 65.87, with a standard deviation of σ = 87.08. You select a random sample of 25 children of daycare age in this same school district. In this sample, you find that the average number of minutes the children spend in active play on weekends is M = 59.28, with a standard deviation of s = 95.79.
The difference between M and mu is due to the sampling error is ______.
Mathematics
1 answer:
alisha [4.7K]3 years ago
8 0

Answer: The difference between M and mu is due to the sampling error is <u>-6.59 .</u>

Step-by-step explanation:

As per given:  

in population, the average number of minutes spent in active play on weekends is μ = 65.87

In sample, the average number of minutes the children spend in active play on weekends is M = 59.28

Now, the difference between M and μ is due to the sampling error is M- μ

= 59.28  - 65.87

= -6.59

So, the difference between M and mu is due to the sampling error is <u>-6.59 .</u>

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Step-by-step explanation:

Let

x ----> number of nickels

y ----> number of dimes

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1\ nickel=\$0.05

1\ dime=\$0.10

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0.05x+0.10y=9.45 ------> equation A

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<em>Solve the system by substitution</em>

we have

0.05x+0.10y=9.45 ------> equation A

0.05x+0.20y=16.65 ------> equation B

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substitute equation C in equation D

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Solve the system by graphing

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