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2 years ago

In 2016, the city of Rio de Janeiro had a population density of 5377 people/km2

Mathematics

1 answer:

Kryger [21]2 years ago

3 0

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Nine friends just graduated from college with business degrees and are ready to start their own businesses. The presentations be

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3 years ago

Read 2 more answers

Look at the picture. Solve for y

Mumz [18]

Answer:

y = 5

Step-by-step explanation:

<u>Equation:</u>

m<DGF = m<DGE + m<EGF

<u>Given:</u>

m<DGF = 12y - 5

m<DGE = 5y + 6

m<EGF = 24

<u />

m<DGF = m<DGE + m<EGF

12y - 5 = 5y + 6 + 24

12y - 5 = 5y + 30

12y - 5y = 30 + 5

7y = 35

y = 5

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3 years ago

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a random

Zielflug [23.3K]

Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .

Step-by-step explanation:

We are given that ,

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

i.e. $\sigma= 12$

Also, it is given that , Sample mean $\overline{x}=98$ having sample size : n= 36

For 95% confidence ,

Significance level : $\alpha=1-0.95=0.05$

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

$z_{\alpha/2}=z_{0.025}=1.96$

We know that the confidence interval for unknown population mean $(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

$\sigma$ = Population standard deviation

$z_{\alpha/2}$ = Critical z-value.

Substitute all the given values, then the required confidence interval will be :

$98\pm (1.96)\dfrac{12}{\sqrt{36}}$

$=98\pm (1.96)\dfrac{12}{6}$

$=98\pm (1.96)(2)$

$=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)$

Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .

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The total price of a printing job at Incredible Invites includes the cost per invitation plus a one-time set-

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Step-by-step explanation:

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