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Korvikt [17]
3 years ago
10

If f(x)=9−4 |x+1|, find f(−5). −15 −7 7 23

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

If f(x) = 9–4 |+1|, find f(-5)?

Step-by-step explanation:

15

-7

7

23

ANSWER 23

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Find the area of the trapezoid with a base of 19 inches, height 12.6 inches and base 29.2 inches
aksik [14]
Minor base: b=19 inches
Height: h=12.6 inches
Major base: B=29.2 inches

Area of the trapezoid: A
A=(b+B)h/2
Replacing the values:
A=(19 inches + 29.2 inches) (12.6 inches) / 2
A=(48.2 inches) (12.6 inches) / 2
A= (607.32 inches^2 ) /2
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Answer: The area of the trapezoid is 303.66 square inches
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What is the combined lenght of all buttons longer than 1 inch?
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Answer:

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Step-by-step explanation:

1: 1 1/4 + 1 1/4

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3:  2 2/4

Answer:  2 2/4

<em><u>Hope this helps.</u></em>

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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
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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}}&#10;And in order to find the sample mean we just need to use this formula:&#10;[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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Answer:

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Step-by-step explanation:

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