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mr_godi [17]
3 years ago
15

Jack rolled a die 60 times and recorded the results in the table. What could Jack expect to happen to the probability of getting

a four if he increased the number of trials to 450?
Mathematics
2 answers:
Aliun [14]3 years ago
5 0
The experimental probability will get closer to the theoretical probability.

At any time during an experiment, you can find the probability of an event. The more trials that you have, the closer you generally get to the theoretical value.
blsea [12.9K]3 years ago
3 0

Answer:

The probability of getting a certain number on a die is independent on the number of trials. It is 1/6.

Step-by-step explanation:

I interpret this question as: what is the probability of getting a 4 on the 450th try? The answer is that precedent tries do not influence the next one. So the probability of getting a four is 1/6 because every roll of the die is an independent event.

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The area of a triangle is 56.4cm and the base is 12cm.What is the height of the triangle?
ale4655 [162]

area of a triangle = 1/2 base x height

56.4 = 1/2 x 12 x height

56.4 = 6 x height

height = 56.4 /6 = 9.4cm


7 0
4 years ago
What is the value of 11−3m , when m = 2? (Please do it step by step because I dont know how to do this ;[ )
STatiana [176]
Substitute the variable for the value given.

11 - 3m becomes 11 - 3(2)

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3 years ago
Translate the algebraic expression shown below into a verbal expression.
yaroslaw [1]
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3 years ago
The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand
NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in  foreign language=0.20

The probability than an American CEO can not transact business in foreign language=1-0.20=0.80

Total number of American CEOs  chosen=12

a. The probability that none can transact business in a foreign language=12C_0(0.20)^0(0.80)^{12}

Using binomial theorem nC_r(1-p)^{n-r}p^r

The probability that none can transact business in a foreign language=\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}

b.The probability that at least two can transact business in a foreign language=1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))

c.The probability that all 12 can transact business in a foreign language=12C_{12}(0.8)^0(0.2)^{12}

The probability that all 12 can transact business in a foreign language=\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}

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3 years ago
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