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MrMuchimi
2 years ago
8

|-4w+2|=14 the bars mean absolute value

Mathematics
2 answers:
Pie2 years ago
5 0

Answer:

If you are looking for what w equals, the w would equal 3

Step-by-step explanation:

You would take the absolute value which is how far away a value is from zero.

So you would get 4w+2=14

Then you would bring the positive 2 to the other side making it negative

now we have 4w=14-2 which is the same as 4w=12

and then, you would divide 12 by 4 and then you would get 3

so

w=3

Ostrovityanka [42]2 years ago
4 0

Answer:

w = 3

Step-by-step explanation:

If we simplify the equation, it will result in 4w + 2 = 14. If we subtract both sides of the equation by 2 to isolate the variable, we will get 4w = 12. To get the value of w, we then have to divide both sides by 4, which gives us our answer that w = 3.

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The answer is C

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ziro4ka [17]

Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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3 years ago
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cos(60) - sin(60) = cos^2(0) + tan(0)

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