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1 year ago

5 pointsWhich is the image of vertex Lafter the triangle is rotated 180 degrees about the origin?

Mathematics

1 answer:

Anastasy [175]1 year ago

6 0

The image of vertex L after the triangle is rotated 180 degrees about the origin= (-2, -1).

Option B.

First let's stae the the original location of the figure on the graph.

From the graph:

L (2, 1)

W(1, -3)

O (-1, -2)

A rotation of 180 degrees about the origin will change the x and y axis of each location to the opposite of those numbers;

Positive numbers becomes negative and negative numbers becomes positive.

The new location for L = L'

L' = (-2, -1)

The new location for W = W'

W' = (-1, 3)

The new location for O = O'

O' = (1, 2)

Hence, the image of vertex L after the triangle is rotated 180 degrees about the origin= (-2, -1).

Option B.

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Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

Write the equation of the line that passes through the points (-4,8) and (-2, -1).

IRINA_888 [86]

Answer:

y=-9/2x-10

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-1-8)/(-2-(-4))

m=-9/(-2+4)

m=-9/2

y-y1=m(x-x1)