Can yall help me please?

4x – 9y = 7 –2x 3y = 4 what number would you multiply the second equation by in order to eliminate the x-terms when adding to th

steposvetlana [31]

Multiply equation II by 2 and then add up the equations.

8 0

3 years ago

Read 2 more answers

Write an equation of the line passing through the points (4,6) and (-2, - 18).

olga_2 [115]

Answer:

y = 4x – 10

Step-by-step explanation:

slope formula: $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

point-slope formula: $y-y_{1}=m\left(x-x_{1}\right)$

Use slope formula and plug in your numbers: $m = \frac{-18-6}{-2-4 }$
Solve for slope $m = 4$
Use the point-slope formula to solve for the equation: $y-6=4\left(x-4\right)$
Solve and find equation: $y=4x-10$

Note: When given two points you only need to pick one for the point-slope formula! Best to choose the positive one so you don't have to deal with the negatives!

Hope this helps! Please give Brainliest!

4 0

2 years ago

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

The cost of safely disposing/A of the toxic chemicals is approximately/B

melomori [17]

The cost of safely disposing of the toxic chemicals is approximately five times what the company paid to purchase it.(Option D)

<h3>What is sentence correction?</h3>

Sentence correction or sentence improvement is a type of grammatical practice where a sentence is given with a word or a phrase that requires grammatical changes or improvement.

Now,

Here, we are talking about "toxic chemicals", that is "plural" but in the Part D of the sentence, "it" refers to the singular chemical.

Hence, the correction needs to be made at Option D; the correction would be: "paid to purchase them."

To learn more about sentence correction, refer to the link: brainly.com/question/14632568

#SPJ4

3 0

1 year ago

Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh

s344n2d4d5 [400]

Answer: The point estimate for the mean weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample $(\overline{x})$ = 31.2 pounds .

Step-by-step explanation:

The weights of spawning Chinook salmon in the Columbia river are normally distributed.

Given : The mean weight from your sample is 31.2 pounds with a standard deviation of 4.4 pounds.

The point estimate for the mean $(\mu)$ weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample $(\overline{x})$ i.e. is 31.2 pounds because the best point estimate for population mean is the sample mean. [in normal distribution]

8 0

3 years ago