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2 years ago

-7w + 2(w + 8) = -24 Can you help me find w?

1 answer:

6 0

-7w + 2(w + 8) = -24

First, expand your equation.
$-7w + 2w + 16 = -24$
Second, add '-7w + 2w' to get '5w'.
$-5w + 16 = -24$
Third, take away '16' from both sides.
$-5w = -24 - 16$
Fourth, subtract '-24 - 16' to get '-40'.
$-5w = -40$
Fifth, divide both sides by '-5' and change it into a positive.
$w = \frac{40}{5}$
Sixth, how many times does 5 go into 40? 40 ÷ 5 = 8.
$w = 8$

Answer: w = 8

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A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w

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The hypothesis is:

H₀: $p_{X}-p_{Y}=0$ .

Hₐ: $p_{X}-p_{Y}$ .

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Let X = number of men who exercise regularly and Y = number of women who exercise regularly.

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$n_{X}=150\\X=88\\n_{Y}=200\\Y=130$

Compute the sample proportion of men and women who exercise regularly as follows:

$\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587$

$\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65$

The random variable X follows a Binomial distribution with parameters n = 150 and $\hat p_{X}=0.587$ .

The random variable Y also follows a Binomial distribution with parameters n = 200 and $\hat p_{Y}=0.65$ .

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\hat p=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion z-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

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H₀: The proportion of women is same as men who exercise regularly, i.e. $p_{X}-p_{Y}=0$ .

Hₐ: The proportion of women is more than men who exercise regularly, i.e. $p_{X}-p_{Y}$ .

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