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sp2606 [1]
3 years ago
13

Can anyone tell me what this is, what unit it is, and how to solve it... Please and thanks. Problem is in the picture.

Mathematics
2 answers:
OlgaM077 [116]3 years ago
6 0

Answer:

10

p + Q= square root of p square + Q Square

Step-by-step explanation:

p(+)q=sqr(p^2+q^2)

8(+)6=sqr(8^2+6^2)

The square root of(64+36)

The square root of(100)

=10

Hope this helped! pls mark brainliest

Vladimir [108]3 years ago
3 0

Answer:

<h3>p(+)q=SQR(p^2+q^2)</h3>

8(+)6=SQR(8^2+6^2)

SQR(64+36)

SQR(100)

=10

Step-by-step explanation:

p + Q is equal to square root of p square + Q Square

sqr means square root

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When consumers apply for credit, their credit is rated using FICO (Fair, Isaac, and Company) scores. Credit ratings are given be
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Answer:

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

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.

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The data is:

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In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

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

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is \bar X= 660.313

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. The degrees of freedom are given by:

df=n-1=16-1=15

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for t_{\alpha/2}=-2.95

"=T.INV(1-0.005,15)" for t_{1-\alpha/2}=2.95

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

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And if we find the limits we got:

660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588  

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]  

So the 99% confidence interval would be given by (589.588;731.038)

Part b

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