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3 years ago

Express as a single power of 2

Mathematics

1 answer:

den301095 [7]3 years ago

6 0

When you multiply powers, they add, and when you divide them they subtract.
So I would first add the 2^a + 2^b +2^c to get 2^(a+b+c)
Then, divide by 2^(a+b). Because when you divide powers they subtract, you will be taking away the (a+b) from the (a+b+c) and you will be left with c on its own.
The answer is 2^c

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According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age grou

Murljashka [212]

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

$p=X/n=35/140=0.25$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239$

We calculate the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198$

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

$P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183$

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

$p=X/n=21/140=0.15$

We can calculate then the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906$

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

$P(X$

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231$

We can calculate the z-scores for p1=0.18 and p2=0.25:

$z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651$

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

$P=P(0.18$

5 0

4 years ago

If 65 percent of the population of a large community is in favor of a proposed rise in school taxes, approximate the probability

xz_007 [3.2K]

Answer: a) 0.99917 b) 0.7054 c) 0.9820

Step-by-step explanation:

Let X denotes the number of people are in favor of a proposed rise in school taxes.

Era re given that p(probability that a person is in favor of a proposed rise )= 0.65

Sample size : n= 100

Mean= np = 100(0.65)=65

Standard deviation: $\sigma=\sqrt{np(1-p)}\\\\=\sqrt{100(0.65)(0.35)}\\\\=4.76969600708\approx4.77$

a) Now, the probability that a random sample of 100 people will contain at least 50 who are in favor of the proposition :

$P(x\geq50)=P(\dfrac{x-\mu}{\sigma}\geq\dfrac{50-65}{4.77})\\\\=P(z\geq-3.144)\\\\=P(z-z)=P(Z$

b) between 60 and 70 inclusive who are in favor;

$P(60$

c) fewer than 75 in favor.

$P(x$

3 0

3 years ago