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3 years ago

Write the equation of the line that is perpendicular to the liney=-1x/ 3 + 2 and goes through the point (5, 6)

Mathematics

1 answer:

KIM [24]3 years ago

3 0

Answer:

y = 3x-9

Step-by-step explanation:

y = -1/3 x +2

The slope is -1/3

We want a line perpendicular to it

Perpendicular lines have negative reciprocal slopes

The new slope is -(-3/1) = 3

We have a slope m=3 and a point (5,6)

We will use point slope form to find the line

y-y-1 = m(x-x1)

y-6 = 3(x-5)

In slope intercept form

Distribute

y-6 = 3x-15

Add 6 to each side

y-6+6 = 3x-15+6

y = 3x-9

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2(4x-3)>-3(3x)+5x help

Roman55 [17]

Answer:

$x>1/2$

Step-by-step explanation:

$2(4x-3)>-3(3x)+5x$

Expand the brackets.

$8x-6>-9x+5x$

Add -9x and 5x

$8x-6>-4x$

Add 4x and 6x on both sides.

$8x+4x>6$

Add the like terms.

$12x>6$

Divide 12 into both sides.

$12x/12>6/12$

$x>1/2$

3 0

2 years ago

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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

2 years ago

Solve the equation 4y – 3 = 1 for y.<br> A.-1<br> B.0<br> C.1<br> D.2

zimovet [89]

Answer:

Step-by-step explanation:

4 times 1 is 4 and 4-3 is 1

8 0

3 years ago

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Find the slope of the line

Alinara [238K]