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muminat
3 years ago
6

In a certain Algebra 2 class of 29 students, 13 of them play basketball and 7 of them play baseball. There are 4 students who pl

ay both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?
Mathematics
1 answer:
Inessa [10]3 years ago
3 0

Answer:

16/29

Step-by-step explanation:

P(A∪B) = P(A) + P(B) - P(A∩B)

P(basketball or baseball) = P(basketball) + P(baseball) - P(both)

= (13/29) + (7/29) - (4/29)

= 16/29

The probability that a randomly chosen student plays either sport is 16/29.

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3p+2.0=31.1
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3 years ago
Rasida ran 5 more miles than 2 times the number of miles Diallo ran this week. Rasida ran 27 miles this week.
Sholpan [36]

Answer:

11

Step-by-step explanation:

Rasida ran 27 miles total.  she ran 5 more miles than double what Diallo ran.

27-5=22

Now divide 22 by 2 to get how many miles Diallo ran.

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3 years ago
A college committee consists of 3 freshmen, 4 sophomores, 5 juniors, and 2 seniors. A subcommittee of 4 consisting of 1 person f
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Answer:

120

Step-by-step explanation:

The subcommittee needs to contain 1 out of 3 freshmen, 1 out of 4 sophomores, 1 out of 5 juniors and 1 out of 2 seniors. Assuming order is not important, the number of possibilities to pick the subcommittee members is given by:

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5 0
3 years ago
A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean     x = 31,5

Sample standard deviation   s₂  = 8,35

Sample size        n₂ = 11

Ti-84 Calculator

Sample mean     y = 46,2

Sample standard deviation   s₁  = 9,99

Sample size        n₁ = 12

t(s)  = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s)  = 12,9 / √(99,8/12) + (69,72/11)

t(s)  = 12,9 / √8,32 + 6,34

t(s)  = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis               H₀      y - x > 1,80

Altenative Hypothesis       Hₐ      y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c)   critical value for

t(c) α; n₁ +n₂-2            df = 12 +11 -2     df  = 21

If we choose  CI = 95 %    then  α = 5 %   α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

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