Hey guys since nobody ever helps me and my mom doesn’t I need help on this assiment thanks :)

Makovka662 [10]

$\frac{ 6\sqrt{2}}{ \sqrt{3} }$

Multiply the square root of 3 by the square root of 2 and itself.
The 3 cancels itself out and then you get A

6 0

3 years ago

What is the solution to the system of equations graphed below?

umka2103 [35]

Answer:

a

Step-by-step explanation:

the two lines intercept at 1,-2

6 0

2 years ago

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PLEASE HELP ME!!!!!

Irina18 [472]

Answer:

f(6) = 1

Step-by-step explanation:

Function:

f(x) = -2/3x + 5

f(6) = -2/3(6) + 5

f(6) = -4 + 5

f(6) = 1

4 0

3 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

4 0

3 years ago

Plz help. pl and thank u

Verizon [17]

Answer: sorry i cant read it its to small

Step-by-step explanation:

6 0

3 years ago

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