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3 years ago

Help pleasseee •-•

Mathematics

1 answer:

Dafna1 [17]3 years ago

3 0

The picture below shows the box plot.

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Suppose that two balanced, six sided dice are tossed repeatedly and the sum of the two uppermost faces is determined on each tos

Simora [160]

Answer:

$\frac{(2/36)}{(1-(28/36))} = 1/4$

Step-by-step explanation:

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2 years ago

13.62<br> Round to the<br> nearest whole<br> number.

julsineya [31]

Answer:

14 would be your answer :)

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2 years ago

I am half way between two consecutive numbers. The product of the two numbers is 650. what number am I ?

elena-14-01-66 [18.8K]

Answer: 325

650/2=325

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2 years ago

arius flipped a coin 500 times and it landed heads up 300 times. What is the relative frequency of the coin landing heads up bas

guajiro [1.7K]

Answer:

3/5

Step-by-step explanation:

500 time number of coin flipped =500

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relative frequency =300/500

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3 years ago

The probability that a certain hockey team will win any given game is 0.3723 based on their 13 year win history of 385 wins out

Whitepunk [10]

Answer:

0.1505 = 15.05% probability that the hockey team wins 6 games in November

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the team wins, or it does not. The probability of winning a game is independent of winning other games. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that a certain hockey team will win any given game is 0.3723

So $p = 0.3723$

12 games in November

So $n = 12$

What is the probability that the hockey team wins 6 games in November?

This is $P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{12,6}.(0.3723)^{6}.(1-0.3723)^{6} = 0.1505$

0.1505 = 15.05% probability that the hockey team wins 6 games in November

4 0

3 years ago