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3 years ago

9 + 1.34 + .5 (3.50 +1.74) How would you simplify the expression? Explain your steps.

1 answer:

5 0

1. Using PEMDAS, I would first solve the equation inside the parenthesis
(3.50+1.74)= 5.24
Now the equation is:
9+1.34+.5(5.24)
2. Multiply .5(5.24)
9+1.34+2.62=
12.96

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3 years ago

The negative effects of ambient air pollution on children's lung function has been well established, but less research is availa

Helen [10]

Answer:

Step-by-step explanation:

Hello!

The objective is to estimate the population mean of the lung-capacity index (FEV1) the forced volume (in ml) of air that is exhaled in 1 second of children that came from households where coal is used for coal or heating or both. The researcher claims that the use of coal has a negative impact (reduces) on the mean FEV1.

A sample of 512 children that came from households where coal is uses was taken, they obtained a sample mean of X[bar]= 1429ml and a standard deviation of S= 327ml

The study variable is X: the forced volume (in ml) of air that is exhaled in 1 second of a child that lives in a household where coal is used.

There is no information about the distribution of the variable, but the sample size is large enough to be valid to apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

Using this approximation you can estimate the population mean using an approximate Z and 1 - α: 0.95:

X[bar] ± $Z_{1-\alpha /2}$ * (S/√n)

$Z_{1-\alpha /2}= Z_{0.975}= 1.965$

[1426 ± 1.965 * (327/√512)]

[1397.60; 1454.40]ml

Using a confidence level of 95% you'd expect that the interval [1397.60; 1454.40] contains the true average of the lung-capacity index of Chinese children that live in households where coal is used for cooking, hearing or both.

Now using a value of S= 330 and a CI of 95% you need to calculate the sample size required to get an interval amplitude of 50ml

To calculate the sample size given a determined confidence level and amplitude the best is to use the semiamplitude or margin of error (d) of the interval.

The semiamplitude is half of the amplitude. The general structure of a confidence level for the mean is "point estimate" ± "margin fo error"

Then the formula for the margin of error is:

d= $Z_{1-\alpha /2}$ * (S/√n)

If a= 50ml then d= 50/2=25ml

$Z_{1-\alpha /2}= Z_{0.975}= 1.965$

(d* $Z_{1-\alpha /2}$ )= (S/√n)

√n*(d* $Z_{1-\alpha /2}$ )= S

n= [S/(d* $Z_{1-\alpha /2}$ )]²

n= [330/(25*1.965)]²

n= 37.29 ≅ 38

You need to take a sample of 38 childre.

I hope it helps!

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3 years ago