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GREYUIT [131]
3 years ago
11

A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The owner of the store has d

etermined that home delivery will be successful only if the average time spent on a delivery does not exceed 38 minutes. The owner has randomly selected 27 customers and delivered pizzas to their homes. What hypotheses should the owner test to demonstrate that the pizza delivery will not be successful
Mathematics
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

Read step by step explanation

Step-by-step explanation:

The owner already knows that the limit for the average time delivered pizzas is 38 minutes. So we conclude

1.-The resulting mean from sample data ( x ) ( 27 customers) need to be smaller than 38 minutes, any value of sample above 38 minutes means more time for the delivery action and will indicate a failure for the future project

2.-As sample size is smaller than 30 the test has to be  t-student one tail test to the left

Test hypothesis

Null hypothesis                       H₀                  x  = 38

Alternative hypothesis           Hₐ                  x  <  38

We should test at a significance level α  = 0,05       (α = 5%)

If the result of the test is to accept  H₀  delivery project won´t be implemented, if on the other hand, H₀ is rejected then in the condition of the alternative hypothesis we accept Hₐ the sample indicates that we have a smaller average time than 38 minutes.

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In a population of 10,000, there are 5000 nonsmokers, 2500 smokers of one pack or less per day, and 2500 smokers of more than on
Kazeer [188]

Answer:

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

Step-by-step explanation:

We have to write the transition matrix M for the population.

We have three states (nonsmokers, smokers of one pack and smokers of more than one pack), so we will have a 3x3 transition matrix.

We can write the transition matrix, in which the rows are the actual state and the columns are the future state.

- There is an 8% probability that a nonsmoker will begin smoking a pack or less per day, and a 2% probability that a nonsmoker will begin smoking more than a pack per day. <em>Then, the probability of staying in the same state is 90%.</em>

-  For smokers who smoke a pack or less per day, there is a 10% probability of quitting and a 10% probability of increasing to more than a pack per day. <em>Then, the probability of staying in the same state is 80%.</em>

- For smokers who smoke more than a pack per day, there is an 8% probability of quitting and a 10% probability of dropping to a pack or less per day. <em>Then, the probability of staying in the same state is 82%.</em>

<em />

The transition matrix becomes:

\begin{vmatrix} &NS&P1&PM\\NS&  0.90&0.08&0.02 \\  P1&0.10&0.80 &0.10 \\  PM& 0.08 &0.10&0.82 \end{vmatrix}

The actual state matrix is

\left[\begin{array}{ccc}5,000&2,500&2,500\end{array}\right]

We can calculate the next month state by multupling the actual state matrix and the transition matrix:

\left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4950&2650&2400\end{array}\right]

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

To calculate the the state for the second month, we us the state of the first of the month and multiply it one time by the transition matrix:

\left[\begin{array}{ccc}4950&2650&2400\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4912&2756&2332\end{array}\right]

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

If we repeat this multiplication 12 times from the actual state (or 10 times from the two-months state), we will get the state a year from now:

\left( \left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] \right)^{12} =\left[\begin{array}{ccc}4792.63&3005.44&2201.93\end{array}\right]

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

3 0
4 years ago
2x - 11 = -8x + 14<br> Simplify
Sauron [17]

Answer:

x = 2.5

Step-by-step explanation:

2x - 11 = -8x + 14

Add 8x to both sides

10x - 11 = 14

Add 11 to both sides

10x = 25

Divide both sides by 10

x = 2.5

7 0
3 years ago
Read 2 more answers
Which equation is satisfied by all three of the plotted points?
Elan Coil [88]

Answer:

Use desmos graphing calculator plug in your equations and see which one has all points.You can use desmos for a bunch of stuff too

Step-by-step explanation:its C though

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8 0
3 years ago
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In attempting to solve the system of equations y 3x 2 and 6x 2y 4, John graphed the two equations on his calculator. Because he
andreyandreev [35.5K]
No, he is not correct. The lines are the same, so there are an infinite number of solutions.

When two lines are parallel, then the solution is empty set.
4 0
3 years ago
Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = xe−x(1 + y) x ≥ 0 an
Yuri [45]

Answer:

For the given explanation we see that two life times are not independent

Step-by-step explanation:

probability for X (for x≥ 0)

\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dy

-e^-^x\int\limits^\infty_0 {de^-^x^y} \,

e⁻ˣ

Probability for X exceed 3

= \int\limits^\infty_3 {f(x)dx

= \int\limits^\infty_3 {e^-^3 dx

= e^-^3

probabilty for y≥ 0 is

\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dx \\

\int\limits^\infty_0{-x/1+yd(e^-^x^(^1^+^y^))} \, =1/(1+y)² \\

3 0
3 years ago
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