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adoni [48]
2 years ago
8

How do i solve 3v + 1 + 7v = -12

Mathematics
2 answers:
Pie2 years ago
6 0

Answer:

Answer:

v=-13/10

Step-by-step explanation:

Our equation: 3v+1+7v=-12

So, lets solve this step by step:

Step #1 - We first need to combine like terms. This means to combine the vs together. Once we combine them, we get:

10v+1=-12

Step #2 - Now we iscolate the variable(which in this case is v). To do this we subtract the 1 to the other side of the equation. Once we subtract the 1 to the -12, we get:

10v=-13

Step #3 - Finally, we must get rid of the coefficent. To do this, we divide the 10 in front of the v by 10. What we do on one side, we must do on the other, which means we must also divide the -13 by 10 as well. Now we get our answer:

v = -\frac{13}{10}

Hope this helps! :3

riadik2000 [5.3K]2 years ago
4 0

Answer:

-1.3

Step-by-step explanation:

3v+1+7v=-12 equation

10v+1=-12 combine like terms

10v=-13 subtract 1 from both sides

v=-1.3 divide by 10

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GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m
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Answer:

(a) Null Hypothesis, H_0 : \mu \leq 3.50 mg  

     Alternate Hypothesis, H_A : \mu > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average mg of mercury in compact fluorescent bulbs.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, H_A : \mu > 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                          T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean mg of mercury = 3.59

            \sigma = population standard deviation = 0.18 mg

            n = sample of bulbs = 25

So, <em><u>test statistics</u></em>  =  \frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }

                               =  2.50

The value of z test statistics is 2.50.

<u>Now, P-value of the test statistics is given by;</u>

         P-value = P(Z > 2.50) = 1 - P(Z \leq 2.50)

                                             = 1 - 0.9938 = 0.0062

<em />

<em>Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

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