<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B3%7D%20-3%3D%5Cfrac%7Bx%7D%7B9%7D%20%2B3" id="TexFormula1" title="\frac{x}{3}

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nordsb [41]

3 years ago

-3=\frac{x}{9} +3" alt="\frac{x}{3} -3=\frac{x}{9} +3" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Tanya [424]3 years ago

7 0

The answer is : x=27

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Jack is a college athlete who has been weighing himself weekly on the same scale in the athletic center for the past few years.

Radda [10]

Answer:

a) Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=\pm 1.64$

b) $\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{9}}=1$

c) $\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And the margin of error is:

$ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.64$

And then the confidence interval is be given by:

$200-1.64 = 198.36$

$200+1.64 = 201.64$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X= 200$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=3$ represent the population standard deviation

n=9 represent the sample size

Part a

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=\pm 1.64$

Part b

The distribution for the sample mean is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{9}}=1$

Part c

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And the margin of error is:

$ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.64$

And then the confidence interval is be given by:

$200-1.64 = 198.36$

$200+1.64 = 201.64$

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3 years ago

Corsica fills a bucket with water at a constant rate. The equation below represents a proportional relationship between the ounc

alexandr1967 [171]

Answer:

Step-by-step explanation:

6 0

3 years ago

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A researcher drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 209 times. The observed freque

ddd [48]

Answer:

Step-by-step explanation:

Given that the observed frequencies for the outcomes as follows:

To check this we can use chi square goodness of fit test.

$H_0: equally likely\\H_a: not equally likely$

(Two tailed test at 5% significance level)

Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E

Df = 6-1 =5

Outcome Frequency Expected frequency (Obs-exp)^2/Exp

1 36 34.83333333 0.03907496

2 30 34.83333333 0.670653907

3 41 34.83333333 1.091706539

4 40 34.83333333 0.766347687

5 23 34.83333333 4.019936204

6 39 34.83333333 0.498405104

209 209 7.086124402

p value =0.214

Since p >alpha, we accept null hypothesis

It appears that the loaded die does not behave differently than a fair die at 5% level of significance

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The slope in the simplest form

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Y=5/5x+ 5 or y=1 x + 5

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Write each percent as a fraction for 88%

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88 = 88/100 or 22/25

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