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erma4kov [3.2K]
3 years ago
9

Hi, I've recently started doing proofs in math, and I've gotten through most of it, but I am still absolutely clueless when it c

omes to these questions; could someone please help and explain to me how this works??
Question 1: What is the reason for #4?

Given: 4x + 1 = 6x - 2, Prove: x = 3/2

(1) 4x + 1 = 6x - 2

(2) 4x + 3 = 6x

(3) 3 = 2x

(4) 3/2 = x ?

(5) x = 3/2

Question 2:

What is the reason for #3?

Given: 3x - 4 = 2(x + 8), Prove: x = 20

(1) 3x - 4 = 2(x + 8)

(2) 3x - 4 = 2x + 16

(3) 3x = 2x + 20 ?

(4) x = 20

PLEASE HELP I AM DESPERATE TO UNDERSTAND!!
Mathematics
1 answer:
Licemer1 [7]3 years ago
3 0
<span>4x + 1 = 6x - 2, Prove: x = 3/2  
-4x     =-4x
1         =2x-2
-1       =-1
0       =2x-3
0       =2x+(-3)
2/3=2/3 

</span><span>3x - 4 = 2(x + 8), Prove: x = 20 
3x- 4  =2x+16
+4      =+4
3x      =2x+20
-2x     =-2x
1x       =20
</span><span>

</span>
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Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let p_1 = <u><em>proportion of Illinois high school freshmen who have used anabolic steroids.</em></u>

p_2 = <u><em>proportion of Illinois high school seniors who have used anabolic steroids.</em></u>

SO, Null Hypothesis, H_0 : p_1=p_2      {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, H_A : p_1\neq p_2      {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here <u>Two-sample z test for</u> <u>proportions</u>;

                        T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of high school freshmen who have used anabolic steroids = \frac{34}{1679} = 0.0203

\hat p_2 = sample proportion of high school seniors who have used anabolic steroids = \frac{24}{1366} = 0.0176

n_1 = sample of high school freshmen = 1679

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The value of z test statistics is 0.545.

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Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

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