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Anuta_ua [19.1K]
3 years ago
11

If two dice are rolled one time, find the probability of getting: a sum grater than 10\

Mathematics
1 answer:
lilavasa [31]3 years ago
4 0
If two dices are rolled once, you would have a total number of 6*6=36 possibilities.
To get a sum greater than 10:
A) one of your dices could be showing a 5, the other a 6. That’s two possibilities. Dice A being the 5 or dice B being the 5.
B) both of your dices could be showing 6s.
That’s one possibility.
So your overall possibility to get a sum greater than 10 is (1+2)/36 3/36=1/12
One twelfth.
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Coordinates: (0,0) ; (1,5) ; (2,10)

Step-by-step explanation:

x  |  y

0  0

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A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The owner of the store has d
Alinara [238K]

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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3 years ago
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8 0
3 years ago
Read 2 more answers
The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they
sukhopar [10]

Answer:

From the result of the one-tailed z-test, we can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Step-by-step explanation:

Sample size = 50

Mean waiting time = 13 minutes

Standard deviation = 11 minutes

We need to perform one hypothesis test that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

The null hypothesis would be that there is no significant difference.

H₀: μ₀ = 15 minutes

The alternative hypothesis would be that the mean waiting time is indeed significantly less than the 15-minute claim by the taxpayer advocate.

Hₐ: μ₀ < 15 minutes.

This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim). Hence, we can use the z-test

z = (x - μ)/σₓ

σₓ = (σ/√n) = (11/√50) = 1.556

z = (13 - 15)/1.556 = - 1.29

Using the z-table at significance level of 0.05.

p-value = 1 - 0.9115 (from the z-tables)

p-value = 0.0885

Since the p-value is more than the significance level (0.0885 > 0.05), we do not reject the null hypothesis.

Hence, we accept the null hypothesis and can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Hope this Helps!!!

4 0
3 years ago
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