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1 year ago

Select the correct answer.

Mathematics

1 answer:

andrew-mc [135]1 year ago

4 0

The prediction for average temperature is 53.06° F

Given: The equation for line of best fit:

y = -3.32x + 97.05

What is line of best fit?

Line of best fit is the line that is drawn through the points and it minimizes the error between the actual data and the predicted data.

In a regression graph, there are many scattered points and it becomes very difficult to predict a general equal if we focus on the coordinates and relations of each points. The line of best fit, is a line that passes through the center or from a position where the error between the point and the line is the minimum and it becomes easy to predict or a derive an equation or a relation.

What is a response variable?

A response variable is a variable that is dependent on one or more independent or dependent variables for it's value. So let's say in this equation y = mx + c, y is the response variable as on the change of variable x , y will have different values that is y will response differently for different values of x. Hence y is the dependent variable.

What is predictor variable?

A predictor variable is a variable that is independent of the influence from any other variable, and a change in this variable will give different outcomes. It is called predictor because or a change in it's value we can predict an outcome.

Let's solve the question that is given:

Now y is the response variable and x is the predictor variable.

From the given graph, we see

x = inches of rain

y = average temperature

To make the prediction for x = 13.25, we place the value of x in the equation of line of best fit, that is:

y(13.25) = -3.32*13.25 + 97.05 = 53.06

Hence our prediction or the response for average temperature is 53.06° F

Know more on "line of best fit" here: brainly.com/question/14279419

#SPJ1

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$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.