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PtichkaEL [24]
3 years ago
15

company is developing a new​ high-performance wax for cross country ski racing. In order to justify the price marketing​ wants,

the wax needs to be very fast.​ Specifically, the mean time to finish their standard test course should be less than 55 seconds for a former Olympic champion. To test​ it, the champion will ski the course 8 times. The​ champion's times are 59.5​, 64.7​, 46.6​, 54.8​, 47.1​, 46.1​, 52.2​, and 41.6 seconds to complete the test course. Complete parts a and b below. ​a) Should they market the​ wax? Assume the assumptions and conditions for appropriate hypothesis testing are met for the sample. Assume alphaequals0.05. What are the null and alternative​ hypotheses? Choose the correct answer below. A. H0​: muless than55 vs. HA​: muequals55 B. H0​: muequals55 vs. HA​: mugreater than55 C. H0​: mugreater than55 vs. HA​: muequals55 D. H0​: muequals55 vs. HA​: muless than55'
Mathematics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

Correct option: (D) <em>H₀</em>: <em>μ</em> =  55 vs. <em>Hₐ</em>: <em>μ</em> < 55.

Step-by-step explanation:

In this case we need to test whether the new​ high-performance wax price is justifiable or not.

It is provided that the wax needs to be very fast specifically the mean time to finish their standard test course should be less than 55 seconds for a former Olympic champion, for the price to justifiable.

So, to test this a sample of <em>n</em> = 8 champion's times are recorded.

A one-sample <em>t</em>-test will be used to perform the test.

The hypothesis to test this is defined as follows:

<em>H₀</em>: The mean time to finish their standard test course is 55 seconds, i.e. <em>μ</em> =  55.

<em>Hₐ</em>: The mean time to finish their standard test course is less than 55 seconds, i.e. <em>μ</em> < 55.

Thus, the correct option is (D).

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Answer:

43

Step-by-step explanation:

43 is a prime number

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2 years ago
8% as a fraction in simplest form
masha68 [24]

Answer: 2/25

Step-by-step explanation:

8% means 8/100

To reduce to the simplest fraction, divide through with a common factor of 4

Divide numerator by 4 = 2

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Therefore, 8% = 2/25

I hope this helps.

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At a concession stand five hotdogs and two hamburgers cost 8.00 two hotdogs and five hamburgers cost 9.50 find the cost of one h
Svetach [21]
Think of this situation as a system of equations.

let y be the price of hotdogs
let x be the price of hamburgers

5y + 2x = 8
2y + 5x = 9.50

now use your solution method of choice
substitution, elimination, matrices, etc
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Quadrilateral with no parallel sides
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Write an equation in slope-intercept form for the following line:<br><br> (-14,1) and (13,-2)
alex41 [277]

Answer:

\large \boxed{y =  -  \frac{1}{9} x -  \frac{5}{9} }

Step-by-step explanation:

In order to find an equation of a line with two given ordered pairs. We have to find a slope first which we can do by using the formula below.

\large \boxed{m =  \frac{y_2 - y_1}{x_2 - x_1} }

m-term is defined as slope in y = mx+b form which is slope-intercept form.

Now we substitute these ordered pairs (x, y) in the formula.

\large{m =  \frac{1 - ( - 2)}{ -14 - 13} } \\  \large{m =  \frac{1 + 2}{ - 27} } \\  \large{m =  \frac{3}{ - 27}  =  -  \frac{1}{9} }

After we calculate for slope, we substitute m-value in slope-intercept form. The slope-intercept form is

\large \boxed{y = mx + b}

We already know m-value as we substitute it.

\large{y =  -  \frac{1}{9} x  + b}

We are not done yet because we need to find the b-term which is our y-intercept. (Note that m-term is slope while b-term is y-intercept)

We can find the y-intercept by substituting either (-14,1) or (13,-2) in the equation. I will be using (13,-2) to substitute in the equation.

\large{y = -  \frac{1}{9} x + b} \\  \large{ - 2 =  -  \frac{1}{9} (13) + b} \\  \large{ - 2 =  -  \frac{13}{9}  + b} \\  \large{ - 2 +  \frac{13}{9}  = b} \\  \large{ -  \frac{5}{9}  = b}

Finally, we know b-value. Then we substitute it in our equation.

\large{y =  -  \frac{1}{9} x + b} \\  \large{y =  -  \frac{1}{9} x -  \frac{5}{9} }

4 0
2 years ago
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