<img src="https://tex.z-dn.net/?f=343%5E%7Bx-3%7D%3D49%5E%7B2x-2%7D" id="TexFormula1" title="343^{x-3}=49^{2x-2}" alt="343^{x-3}

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Keith_Richards [23]

3 years ago

=49^{2x-2}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

ioda3 years ago

8 0

DONT CLICK ON THAT LINK PLEASE! But the answer is x= -5

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Emma ran 3.5 kilometers while grace ran 380 meters who ran further ???

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The number of students going to the school field trip reduced from 1000 to 950. What is the percent of change? I will help the f

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8_murik_8 [283]

Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

$\bar x = 7.10$ -- sample mean

$\sigma=5$ --- sample standard deviation

$n = 150$ --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

$CI = 90\%$ --- the confidence interval

Calculate the $\alpha$ level

$\alpha = 1 - CI$

$\alpha = 1 - 90\%$

$\alpha = 1 - 0.90$

$\alpha = 0.10$

Divide by 2

$\frac{\alpha}{2} = 0.10/2$

$\frac{\alpha}{2} = 0.05$

Subtract from 1

$1 - \frac{\alpha}{2} = 1 - 0.05$

$1 - \frac{\alpha}{2} = 0.95$

From the z table. the critical value for $1 - \frac{\alpha}{2} = 0.95$ is:

$z = 1.960$

Solving (c): Margin of error

This is calculated as:

$E = z * \frac{\sigma}{\sqrt n}$

$E = 1.960 * \frac{5}{\sqrt {150}}$

$E = 1.960 * \frac{5}{12.25}$

$E = \frac{1.960 *5}{12.25}$

$E = \frac{9.80}{12.25}$

$E = 0.800$

Solving (d): The confidence interval

This is calculated as:

$CI = (\bar x - E, \bar x + E)$

$CI = (7.10 - 0.800, 7.10 + 0.800)$

$CI = (6.3, 7.9)$

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

6 0

3 years ago