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Strike441 [17]
2 years ago
14

Select the non-linear function from the list of the following:

Mathematics
1 answer:
solong [7]2 years ago
6 0

Answer:

c

Step by step explanation: Linear equations don't have exponents, so that makes option c nonlinear

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Rocco has 48.7 cups of flour in his storage unit. He needs 2/18 that amount to use for his homemade volcano. What is the total n
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Step-by-step explanation:

There are 48.7 cups in total

2/18 of 48.7 cups is needed for volcano

Total amount needed for experiments are :

2/18 of 48.7 = 97.4/18 = 5.411 cups

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3 years ago
What is the volume of the prism that can be constructed from this net?
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That would make a rectangular prism with measurements of:

Length = 9
Width = 4
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V = lwh

V = 9 * 4 * 10

V = 360
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Tina has been dieting for a total fo 13 weeks. She lost 3 pounds on the first week of her diet, but gained back a pound on the s
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4 she lost 4 pounds, how do u not know this dood

Step-by-step explanation:


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If Carol hangs a picture
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Answer:

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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
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