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3 years ago

Explain how you can use the picture to show 391 ÷ 23 = 17

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:so the top of the rectangle is the answer so 10 +7

Step-by-step explanation:

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Natali5045456 [20]

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3 years ago

Which angles are corresponding angles?

Mrac [35]

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3 years ago

X^3+13x^2+32x+20<br> please help me factorizing above equation

Mazyrski [523]

$x^3+13x^2+32x+20=x^3+x^2+12x^2+ 12x+20x+20 =\\\\=x^2(x+1)+12x(x + 1 )+20(x+1 )=(x+1)(x^2+12x+20)=\\\\=(x+1)[(x^2+10x+2x+20)]=(x+1)[x(x +10)+2(x+10))]=\\\\=(x+1) (x +2) (x+10)$

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3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

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

4 0

3 years ago

Dan takes an airplane to visit his parents. He pays the airline a total of $450. The cost of the ticket is $315. The airline als

vredina [299]

Answer:

Step-by-step explanation:

Start with the total amount Dan paid and subtract the part of the payment that is for the ticket.

450-315=135

Assuming that all of that $135 was for luggage fees and each pound over the limit costs $15, then the question is, how many times did Dan pay $15 until he'd paid the full $135 luggage fee.

So divide $135 by $15.

135/15=9

So his luggage weighted 59 pounds, which is 9 pounds over the limit.

6 0

2 years ago