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3 years ago

If 21 is added to a number, the result is 41 less than twice the number. Find the number.

Mathematics

1 answer:

TEA [102]3 years ago

5 0

Answer: 62

Step-by-step explanation: Set up your equation

21+x=2x-41 because the number (x) that 21 is added to is 41 less than (so subtraction) twice (so multiply the number x by 2) the number (x).
Get the x's on one side by subtracting the single x
21=x-41
Now add 41 to both sides to completely isolate x
62=x
So, your answer is x=62. The number is 62.

Check your work. 62+21=83. 2(62)= 124. 124-41=83. Both equations equal 83, so the solution is correct.

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Answer:

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3 years ago

A study1 conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and th

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Answer:

a) Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

b) $z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

c) $z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

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Step-by-step explanation:

Information given

$X_{1}=688$ represent the number of men with smartphone

$X_{2}=671$ represent the number of women with smartphone

$n_{1}=989$ sample of men selected

$n_{2}=1012$ sample of women selected

$p_{1}=\frac{688}{989}=0.696$ represent the proportion of men with smartphone

$p_{2}=\frac{671}{1012}=0.663$ represent the proportion of women with smartphone

$\hat p$ represent the pooled estimate of p

z would represent the statistic

$p_v$ represent the value

Part a

We want to test if we have difference in the proportion owning a smartphone between men and women, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

Part b

The statistic for this case is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

Part c

Replacing the info given we got:

$z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

Part d

For this case we see that $\hat p_1 > \hat p_2$ so then the answer for this cae would men

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3 years ago