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Sholpan [36]
3 years ago
6

7 girls audition for 12 roles in a school play. What is the probability that at least 2 of the girls audition for the same part?

Mathematics
1 answer:
Sophie [7]3 years ago
6 0

Answer:

Step-by-step explanation:

This is one minus the probability that all the girls audition for different roles. The total number of ways of assigning roles to the girls is 12^7, because to each of the 7 girls, you have a choice of 12 roles.

Then if each girl is to receive a different role, then there are 12!/5! possibilities for that. If you start assigning roles to the girls, then for the first girl, there are 12 choices, but for the next you have to choose one of the 11 different ones, so 11 for the next, and then one of the 10 remaining for the next etc. etc., and this is 12*11*10*...*6 = 12!/(12-7)! =12!/5!

The probability that a random assignment of one of the 12^7 roles would happen to be one of the 12!/5! roles where each girl has a different role, is

(12!/5!)/12^7 = 12!/(12^7 5!)

Then the probability that two or more girls addition for the same part is the probability that not all the girls are assigned different roles, this is thus:

1 - 12!/(12^7 5!)

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A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might b
vovikov84 [41]

Answer:

p > α

0.7038 > 0.05

Also since F < F critical

0.475 < 3.238

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a difference among the four chemical agents on the strength of a particular type of  cloth.

Step-by-step explanation:

We are given that a chemist wishes to test the effect of four chemical agents on the strength of a particular type of  cloth.

Since we are given data for four independent chemical agents to determine the effect on the strength of a particular type of cloth, therefore, a one-way analysis of variance may be used for the given problem.

ANOVA:

The one-way analysis of variance (ANOVA) may be used to find out whether there is any significant difference between the means of two or more independent categories of data.

Set up hypotheses:

Null hypotheses = H₀: μ₁ = μ₂ = μ₃ = μ₄

Alternate hypotheses = H₁: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄

Set up decision rule:

We Reject H₀ if p ≤ α

OR

We Reject H₀ if F > F critical

ANOVA in Excel:

Step 1:

In the data tab, select data analysis

Step 2:

Select "Anova single factor" from the analysis tools

Step 3:

Select the destination of input data in the "input range"

Step 4:

Select "rows" for the option "Group By"

Step 5:

Tick the option "labels in first row"

Step 6:

Set alpha = 0.05

Step 7:

Select the destination of output data in the "output options"

Conclusion:

Please refer to the attached results.

The p-value is found to be

p = 0.7038

The F value is found to be

F = 0.475

The F critical value is found to be

F critical = 3.238

Since p > α

0.7038 > 0.05

We failed to reject H₀

Also since F < F critical

0.475 < 3.238

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a difference among the four chemical agents on the strength of a particular type of cloth.

6 0
3 years ago
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denis23 [38]

Answer:

x = 29.44

y = 34

Step-by-step explanation:

For this problem, we will simply use trigonometry to find the values of y and x for the given triangle.

The two trigonometric functions we will use are tangent and sine.

The general form for tangent is:

tan (Θ) = opposite / adjacent

The general form for sine is:

sin (Θ) = opposite / hypotenuse

For our given triangle, let's solve for x.

tan (30) = 17 / x

x * tan(30) = (17 / x) * x

x tan(30) = 17

x tan(30) * ( 1 / tan(30) ) = 17 * ( 1 / tan(30) )

x = 17 / tan(30)

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Now, let's solve for y.

sin (30) = 17 / y

y * sin(30) = (17 / y ) * y

y sin (30) = 17

y sin (30) * ( 1 / sin(30) ) = 17 * ( 1 / sin(30) )

y = 17 / sin(30)

y = 34

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x = 29.44

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Cheers.

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Answer:

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Step-by-step explanation:

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