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nalin [4]
3 years ago
10

5x=3x+20 can you answer this and show how you got the answer

Mathematics
2 answers:
charle [14.2K]3 years ago
8 0
5x=3x+20
5x-3x=3x-3x+20
2x=20
2x/2=20/2
X=10   Hope this helped!

-Twix
nikdorinn [45]3 years ago
5 0
5x=3x+20

So what I did was i took the 3x and did this: 5x=3x+20
                                                                     -3x -3x
Then i crossed out both of the 3x and then I took 5 and subtract the 3 which would be 2x=20

Take the 20 and divide by 2 and your answer should be x=10
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pshichka [43]

Answer:

We can see that the number in the middle is 36. Hence the median of the data set is 36 the median of a data set is the number placed at the middle of a data after rearrangement either in ascending order or descending order.

Given the dataset 10, 15, 27, 33, 33, 36, 42, 47, 45, 56, 78

Step-by-step explanation:

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2 years ago
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A county is considering rasing the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 m
DochEvi [55]

Answer:

No, at alpha equals 0.10​, we do not have enough evidence to support the​ county's claim.

Step-by-step explanation:

We are given that a county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 miles per hour.

A random sample of 15 vehicles has a mean speed of 31 miles per hour and a standard deviation of 4.7 miles per hour.

<em><u>Let </u></em>\mu<em><u> = true mean speed of the vehicles.</u></em>

SO, <u>Null Hypothesis</u>, H_0 : \mu \leq  30 miles per hour   {means that the mean speed of vehicles is lesser than or equal to 30 miles per hour}

<u>Alternate Hypothesis,</u> H_A : \mu > 30 miles per hour   {means that the mean speed of vehicles is greater than 30 miles per hour}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

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

where, \bar X = sample mean speed of 15 vehicles = 31 mph

             s = sample standard deviation = 4.7 mph

             n = sample of vehicles = 15

So, <em><u>test statistics</u></em>  =   \frac{31-30}{\frac{4.7}{\sqrt{15} } }  ~ t_1_4

                               =  0.824

<u><em>Hence, the value of test statistics is 0.824.</em></u>

<em />

<em>Now at 0.10 significance level, the t table gives critical value of 1.345 at 14 degree of freedom for right-tailed test. Since our test statistics is less than the critical value of t as 0.824 < 1.345, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>

<em />

Therefore, we conclude that the mean speed of vehicles is lesser than or equal to 30 miles per hour which means that the county's claim is not supported.

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Answer:

Step-by-step explanation:

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