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zavuch27 [327]
3 years ago
14

6(x-2)=-18 solve the equation

Mathematics
2 answers:
ch4aika [34]3 years ago
7 0

Answer:

x = -1

Step-by-step explanation:

First distriute, then add 12 to both sides since we're working with negatives. Then you simplfiy and divide by both sides and you get -1.

Have a great day! :D

djyliett [7]3 years ago
4 0

Answer:

x=-1

Step-by-step explanation:

6x-12=-18

6x=-6

x=-1

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Kisachek [45]

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3 years ago
HELP ASAP!!!!!!!PLEASE SHOW WORK!!!!!! !!!!!
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Answer:

Area = 12.82 miles²

Step-by-step explanation:

Area of a triangle with two adjacent sides and the inscribed angle between these side is given by,

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3 years ago
A library expansion, begun in 2002, was expected to cost $103 million. By 2006, library officials estimate the cost would be $39
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6 0
4 years ago
15-a=20 what value would make this true
sergejj [24]

Answer:

a = -5

Step-by-step explanation:

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Move the constant of 15 to the other side by subtracting it.

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5 0
3 years ago
Read 2 more answers
A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample stand
Tpy6a [65]

Answer:

We conclude that the population mean is different from 10.5.

Step-by-step explanation:

We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.

<em>We have to test the claim that the population mean is 10.5.</em>

Let, NULL HYPOTHESIS, H_0 : \mu = 10.5  {means that the population mean is 10.5}

ALTERNATE HYPOTHESIS, H_a : \mu \neq 10.5  {means that the population mean is different from 10.5}

The test statistics that will be used here is One-sample t-test;

           T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean = 11

            s = sample standard deviation = 2

            \mu = population mean

            n = sample of values = 16

So, <u>test statistics</u> =  \frac{11-10.5}{\frac{2}{\sqrt{16} } } ~ t_1_5

                            = 1

<em>Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.</em>

Therefore, we conclude that the population mean is different from 10.5.

6 0
4 years ago
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