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3 years ago

Use the Exterior Angle Inequality Theorem to list all of the angles that measure less than the measure of Angle 1

Mathematics

1 answer:

kati45 [8]3 years ago

6 0

Answer:

Angle 5

Angle 8

Step-by-step explanation:

$\angle 5 \:\&\: \angle 8$ measure less than the measure of Angle 1.

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Which inequality represents the situation? My dogs weight is greater than 80 pounds.

Nata [24]

Answer:Dog's weight> 80 pounds

Step-by-step explanation:Dog's weight is greater than 80 pounds

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3 years ago

What is the answer to this math problem

vesna_86 [32]

Hello from MrBillDoesMath!

Answer:

1/243

Discussion:

3^(-3) * 3^(-2) => as a^m * a^n = a^(m+n), the exponents add

= 3^ (-3 -2)

= 3^ (-5) =

= 1/ 3^5 => as 3^5 = 243

= 1/243

which is approximately .0041

Thank you,

MrB

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4 years ago

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What is 11/12 - 2/3 <br> Subtracting Proper Fractions

alexgriva [62]

Answer:

0.25

Step-by-step explanation:

11/12 = 0.917

2/3 = 0.667

0.917 - 0.667 = 0.25

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2 years ago

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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

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3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Chuge%5Ccolorbox%7Blightblue%7D%7Bwhat%20%5C%3A%20is%20%5C%3A%20multiverse%20%3F%7D%20" i

almond37 [142]

Answer:

HOPE THIS IS WHAT YOU ARD LOOK FOR

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3 years ago

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