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Sloan [31]
3 years ago
7

Association - Suppose you were to collect data for each pair of variables. You want to make a scatterplot. Which variable would

you use as the explanatory variable and which as the response variable? Why? What would you expect to see in the scatterplot? Discuss the likely direction , form, and strength. a) Apples: weight in grams, weight in pounds b) Apples: diameter (inches), weight (grams) c) College freshmen: height, grade point average d) Gasoline: number of miles you drove since filling up, gas remaining in your tank
Mathematics
2 answers:
Nana76 [90]3 years ago
6 0

Answer:

The answer is B.

Step-by-step explanation:

The question asks us to select an explanatory variable and a response variable which are just different names for independent and dependent variables.

Explanatory or independent variables, as the name suggests, are not affected by any other variable in the equation.

Response or dependent variables change according to the changes in other variables in the equation.

Out of the four options given in the question, the option b should be chosen according to the requirements because diameter of the apple affects its weight but the diameter itself is independent of the weight variable.

The plot should show that the weight changes according to the diameter variable.

I hope this answer helps.

CaHeK987 [17]3 years ago
6 0

Answer: b

Step-by-step explanation:

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Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

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