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Viktor [21]
3 years ago
12

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

t to y from the first equation is substituted into the second equation.
8x – y = –9
4x – 3y = –22

4x – 3(–8x – 9) = –22
4x – 3(8x + 9) = –22
4(8x + 9) – 3y = –22
4(–8x – 9) – 3y = –22

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalent to y from the first equation is substituted into the second equation.

8x – y = –9
4x – 3y = –22

4x – 3(–8x – 9) = –22
4x – 3(8x + 9) = –22
4(8x + 9) – 3y = –22
4(–8x – 9) – 3y = –22
Mathematics
1 answer:
swat323 years ago
3 0
So here is the answer to the given question above. 
If we are to use substitution method in order to solve the following system, the new equation would be <span>4x – 3(8x + 9) = –22, the answer is the second option. 
Here is how I got the answer.
Let's take 8x - y = -22
So y = 8x + 22
Now, substitute the value of y to </span><span>4x – 3y = –22
and we get 4x - 3(8x +9) = -22.
Hope this answer helps. </span>
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A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute
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(a) Null Hypothesis, H_0 : \sigma = 10 beats per minute  

     Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute  

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Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

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Let \sigma = <u><em>population standard deviation for the pulse rates of men</em></u>.

(a) So, Null Hypothesis, H_0 : \sigma = 10 beats per minute      {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute      {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is <u>One-sample chi-square test</u> for standard deviation;

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(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

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