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gayaneshka [121]
3 years ago
5

There are 30 names in a hat. If two names are picked without replacement, which expression shows the probability that Jack and J

ill will be picked?​

Mathematics
1 answer:
Darya [45]3 years ago
7 0

Answer: Choice C)

(1/30)*(1/29)

Explanation:

Jack has a 1/30 chance of being picked since he is 1 person out of 30 total. After his name is picked, and not put back, there are 30-1 = 29 names left. The chances Jill is picked is 1/29. The two fractions are multiplied to get the overall probability both are picked.

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Answer:

72 students

Step-by-step explanation:

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Setting:
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A. Mr. Kent interviewed the 54 students as they are going to leave the school, it is not considered to be a random sample. It is because a random sample is when a set is taken from a population. Mr. Kent interviewed the 54 who are going to leave, meaning, he didn't take a set out of that 54, he took all of them. So it is not a random sample.

b. The question that Mr. Kent asked is considered to be a leading question, so it does not seem biased.

c. If there are 54 respondents.
51 = yes, the rest is no.
= 54 - 51 = 3
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4 years ago
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Between the two of them, they teach 35 yoga classes each week. if erica teaches 13 fewer than twice as many as bo, how many clas
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7 0
3 years ago
Given 3(x-2)=6, then x is equal to: Select one: a. 5 b. 4 c. 2 d. 6
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6 0
4 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

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tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

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or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

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3 years ago
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