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3 years ago

HD Straight line graphs

Mathematics

1 answer:

Taya2010 [7]3 years ago

5 0

Answer:

The equation of the sraight line 3x- y+ 6 =0

Step-by-step explanation:

Explanation:-

Given that the gradient of the function

|gradf| = 3 and point (-1,3)

Given that the slope of the line

m = 3

The equation of the straight line passing through the point(-1,3) and slope m =3

$y -y_{1} = m(x-x_{1} )$

y-3 = 3(x-(-1))

y-3 = 3x+3

3x +3-y+3=0

3x- y+ 6 =0

Final answer:-

The equation of the sraight line 3x- y+ 6 =0

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Step-by-step explanation:

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If you double the lengths of a rectangle what can you say about the new area of it.

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4 years ago

Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbar

Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : $n=45$

$\overline{x}=148.79\text{ lb}$

$\sigma=31.37\text{ lb}$

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = $\mu=148.79\text{ lb}$

b) The confidence interval for the population mean is given by :-

$\mu\ \pm E$ , where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Margin of error : $E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69$

Now, the 90% confidence interval for the population mean will be :-

$148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)$

Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0

3 years ago

Given the matrices: 1 2 A= 1 -1 2 1 1 B= 3 4 Calculate AB: C11 C12 [2.1] х 1 2 3 4 C21 C22 C11 = C12 = -2 C22 - C215 DONE

eduard

Answer:

$\:c_{11}=-2,\:\:\:c_{12}=-2$

$\:c_{21}=5,\:\:\:c_{22}=8$

Step-by-step explanation:

Given the matrices

$A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}$

$B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}$

Calculating AB:

$\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}$

Multiply the rows of the first matrix by the columns of the second matrix

$=\begin{pmatrix}1\cdot \:1+\left(-1\right)\cdot \:3&1\cdot \:2+\left(-1\right)\cdot \:4\\ 2\cdot \:1+1\cdot \:3&2\cdot \:2+1\cdot \:4\end{pmatrix}$

$=\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}$

Hence,

$\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}$

Therefore,

$\:c_{11}=-2,\:\:\:c_{12}=-2$

$\:c_{21}=5,\:\:\:c_{22}=8$

5 0

3 years ago

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