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2 years ago

3(1) to the power of 2

Mathematics

1 answer:

love history [14]2 years ago

7 0

Answer:

Step-by-step explanation:

Raising a number to the second means multiplying that number. 3 * 3 = 9

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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

Helppppppp hugssssssssssssss

BlackZzzverrR [31]

ANSWER: y=x-8

EXPLANATION:

4 0

2 years ago

Read 2 more answers

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi

Margaret [11]

Answer:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=26Kg$

Step-by-step explanation:

Data

Brine a (Ba)

$V_{Ba}=5\frac{Lt}{min}\\ Concentration(Bca)=0.05\frac{Kg}{Lt}$

Brine b (Bb)

$V_{Bb}=10\frac{Lt}{min}\\ Concentration(Bcb)=0.04\frac{Kg}{Lt}$

we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)

, then the amount of salt (y) left in the tank after (t) minutes: $y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}$