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Crank
2 years ago
10

For a specific species of fish in a pond, a wildlife biologist wants to build a regression equation to predict the weight of a f

ish based on its length. The biologist collects a random sample of this species of fish and finds that the lengths vary from 0.75 to 1.35 inches. The biologist uses the data from the sample to create a single linear regression model. Would it be appropriate to use this model to predict the weight of a fish of this species that is 3 inches long
Mathematics
1 answer:
sashaice [31]2 years ago
3 0

Answer: First of all, we will add the options.

A. Yes, because 3 inches falls above the maximum value of lengths in the sample.

B. Yes, because the regression equation is based on a random sample.

C. Yes, because the association between length and weight is positive.

D. No, because 3 inches falls above the maximum value of lengths in the sample.

E. No, because there may not be any 3-inch fish of this species in the pond.

The correct option is D.

Step-by-step explanation: It would not be appropriate to use the model to predict the weight of species that is 3 inches long because 3 inches falls above the maximum value of lengths in the sample.

As we can see from the question, the model only accounts for species that are within the range of 0.75 to 1.35 inches in length, and species smaller or larger than that length have not been taken into consideration. Therefore the model can not be used to predict the weights of fishes not with the range accounted for.

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Lubov Fominskaja [6]
<h2>Explanation:</h2><h2></h2>

We know that the volume of a prism is defined by:

V=lwh \\ \\ \\ Where: \\ \\ l:length \\ \\ w:width \\ \\ h:height \\ \\ \\ l=\frac{d-2}{3d-9} \\ \\ w=\frac{4}{d-4} \\ \\ h=\frac{2d-6}{2d-4}

Substituting values:

V=\left(\frac{d-2}{3d-9}\right)\left(\frac{4}{d-4}\right)\left(\frac{2d-6}{2d-4}\right) \\ \\ \\ Simplifying: \\ \\ V=\frac{d-2}{3d-9}\cdot \frac{4}{d-4}\cdot \frac{d-3}{d-2} \\ \\ V=\frac{\left(d-2\right)\cdot \:4\left(d-3\right)}{\left(3d-9\right)\left(d-4\right)\left(d-2\right)} \\ \\ V=\frac{4\left(d-3\right)}{\left(3d-9\right)\left(d-4\right)} \\ \\ V=\frac{4\left(d-3\right)}{3\left(d-3\right)\left(d-4\right)}

Finally: \\ \\  \boxed{V=\frac{4}{3\left(d-4\right)}}

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2 years ago
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Answer:

Total cost for tiles and paints is $924.  

Step-by-step explanation:

We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.

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Let us find area of walls and ceiling.

\text{Area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)

\text{Area of walls and ceiling}=1200+90+120

\text{Area of walls and ceiling}=1410

Therefore, the area of walls and ceiling is 1410 square meters.

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\text{ The total painting cost}=10*(\frac{1410}{25})

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Therefore, the total painting cost is $564.  

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Answer:

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Step-by-step explanation:

We have been given a data set that represents the ages of 36 executives. We are asked to find the percentile that corresponds to an age of 41 years.

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