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3 years ago

Suppose x=0.5(300-x). what is the value of x?

Mathematics

1 answer:

Sonbull [250]3 years ago

6 0

Expand

x = 150 - 0.5x

Add 0.5x to both sides

x + 0.5x = 150

Simplify x + 0.5x to 1.5x

1.5x = 150

Divide both sides by 1.5

x = 150/1.5

Simplify 150/1.5 to 100

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Elan Coil [88]

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150

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We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$ . And replacing we got:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

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