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makvit [3.9K]
3 years ago
5

How do I solve for X HELP

Mathematics
2 answers:
morpeh [17]3 years ago
7 0

Step-by-step explanation:

here \: is \: your \: solution :  -  \\  \\ both \: angles \: are \: equal \: (vertical \: anges) \\  \\ 26x + 7 = 31x - 13 \\  \\ solving \: the \: equation \:  \\  \\ 26x - 31x =  - 13 - 7 \\  \\  - 5x =   20 \\  \\ x =  -20 \div 5 \\  \\ x =  \:   4 \: degrees  \:  \:  \:  \:(Answer ✓✓✓( \\  \\ \huge\mathcal\blue{happy \: life (. ❛ ᴗ ❛.) }

yarga [219]3 years ago
3 0

Answer:

x = 4 degrees

Step-by-step explanation:

The two angles have the same measurement. They are opposite angles.

26x + 7 = 31x - 13

Group like terms:

26x - 31x = -13 - 7

-5x = -20

x = 4

<em>Check if you want to make sure.</em>

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While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
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We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

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

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

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