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3 years ago

Which of the following matches the graph of f (x) = 2(x – 1)2(x + 1)(x + 2)?

Mathematics

1 answer:

nadya68 [22]3 years ago

6 0

Answer:

where is the rest of the question.......

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What is the slope of the line passing through (16, -2) and (-30, 6)?

Novay_Z [31]

$m=\frac{\Delta y}{\Delta x}=\frac{-2-6}{16-(-30)}=\frac{-8}{46}=\boxed{-\frac{4}{23}}$ .

Hope this helps.

5 0

3 years ago

Read 2 more answers

Help needed fast!!!

kicyunya [14]

Answer:

The answer is A

Step-by-step explanation:

It has to exceed not be same or equal to

6 0

3 years ago

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Drop and Drag <br> A)reflection (in x)<br> B)reflection (in y)<br> C)rotation<br> D)translation

djyliett [7]

Answer:

Step-by-step explanation:

because y is a rotation that rotates up and down like a ramp!

4 0

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

4 years ago

Berniece solved the following equation:

aniked [119]

Answer:

Step 4.

Step-by-step explanation:

To isolate the x variable, she subtracts 8 from both sides making it another subtraction property of equality justification.

5 0

3 years ago