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USPshnik [31]
3 years ago
9

* George scored on average of 80% of three tests.

Mathematics
1 answer:
Zigmanuir [339]3 years ago
6 0

Answer:   100%

Step-by-step explanation:

100 + 80 + 80+ 80 = 340  (total scores of 4 tests)

340 divided by the number of tests 4 = 85

George would need to score 100% on his next test to bring his test average from 80 to 85.

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this triangular prism has a length of 20 centimeters, an end base of 6 centimeters and an end height of 8 centimeters. what is t
pochemuha

Answer:

960

Step-by-step explanation:

LxWxH = v

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7 0
3 years ago
he probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1034
garri49 [273]

Answer:

0.0071 = 0.71% probability that the San Jose Sharks win 9 games in the upcoming month.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the Sharks win, or they do not. The probability of the Sharks winning a game is independent of any other game. This means that the binomial probability distribution is used 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 the San Jose Sharks will win any given game is 0.3694.

This means that p = 0.3694

An upcoming monthly schedule contains 12 games.

This means that n = 12

What is the probability that the San Jose Sharks win 9 games in the upcoming month?

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

P(X = 9) = C_{12,9}.(0.3694)^{9}.(0.6306)^{3} = 0.0071

0.0071 = 0.71% probability that the San Jose Sharks win 9 games in the upcoming month.

6 0
3 years ago
According to the data, the mean quantitative score on a standardized test for female college-bound high school seniors was 500.
Sveta_85 [38]

Answer:

6.68% of the female college-bound high school seniors had scores above 575.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 50

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(scores above 575)

P(x > 575)

P( x > 575) = P( z > \displaystyle\frac{575 - 500}{50}) = P(z > 1.5)

= 1 - P(z \leq 1.5)

Calculation the value from standard normal z table, we have,  

P(x > 575) = 1 - 0.9332= 0.0668 = 6.68\%

6.68% of the female college-bound high school seniors had scores above 575.

7 0
3 years ago
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