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dusya [7]
3 years ago
12

A marathon runner is able to run 15 miles in 3 hours. write an equation that relates the number of miles, y, to the number of ho

urs, x.
Mathematics
1 answer:
timama [110]3 years ago
6 0

Answer:

5y = x

Step-by-step explanation:

15 miles = 3 hours

Divide by 3 on both sides to find how many miles he/she can run in 1 hour.

15 miles = 3 hours ⇒ 5 miles = 1 hour

Therefore, the equation is just 5y = x.

Hope this helps, please mark brainliest and give good reviews thanks :)

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Answer:

6 7/9

Step-by-step explanation:

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3 years ago
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A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an a
Taya2010 [7]

Answer:

(a) Null Hypothesis, H_0 : \mu = 23%    

    Alternate Hypothesis, H_A : \mu \neq 23%

(b) We conclude that the mean percentage is  different from 23 and the manufacturing process will be re-calibrated.

(c) P-value is 0.6%.

Step-by-step explanation:

We are given that a certain manufactured product is supposed to contain 23% potassium by weight.

A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2.

Let \mu = <u><em>mean percentage of potassium by weight.</em></u>

(a) Null Hypothesis, H_0 : \mu = 23%      {means that the mean percentage is equal to 23 and the manufacturing process will not be re-calibrated}

Alternate Hypothesis, H_A : \mu \neq 23%     {means that the mean percentage is  different from 23 and the manufacturing process will be re-calibrated}

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

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

where, \bar X = sample mean percentage = 23.2

            s = sample standard deviation = 0.2

            n = sample of specimens = 10

So, <u><em>the test statistics</em></u>  =  \frac{23.2-23}{\frac{0.2}{\sqrt{10} } }  ~ t_9

                                     =  3.162

The value of t test statistic is 3.162.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical value of -2.262 and 2.262 at 9 degree of freedom for two-tailed test.</u>

(b) Since our test statistic doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <em><u>we reject our null hypothesis</u></em>.

Therefore, we conclude that the mean percentage is  different from 23 and the manufacturing process will be re-calibrated.

(c) The P-value of the test statistics is given by;

            P-value = P( t_9 > 3.162) = <u>0.006 or 0.6%</u>

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