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2 years ago

8

"Woohoo! The school carnival is here! You think about all the fun carnival games that you love to play every year. Then you reme

mber that some of the probability concepts you learned about in math class could help you here. Time to play!"

1 answer:

Lena [83]2 years ago

5 0

Answer:

14

Step-by-step explanation:

You might be interested in

On a multiple-choice test. Abby randomly guesses on all seven questions. Each question

Murrr4er [49]

Answer:

0.173 probability that she gets exactly three questions correct.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either she guesses the correct answer, or she does not. The probability of guessing the correct answer for a question is independent of other questions. 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.

Seven questions:

This means that $n = 7$

Each question has four choices.

Abby guesses, which means that $p = \frac{1}[4} = 0.25$

Find the probability to the nearest thousandth, that Abby gets exactly three questions correct.

This is P(X = 3).

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

$P(X = 3) = C_{7,3}.(0.25)^{3}.(0.75)^{4} = 0.173$

8 0

2 years ago

RSM wants to send four of its 18 Math Challenge teachers to a conference. How many combinations of four teachers include exactly

tatiyna

Answer:

1120 combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan.

Step-by-step explanation:

The order in which the teachers are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

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

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

In this question:

1 from a set of 2(Either Mrs. Vera or Mr. Jan).

3 from a set of 18 - 2 = 16. So

$C_{2,1}C_{16,3} = \frac{2!}{1!1!} \times \frac{16!}{3!13!} = 1120$

1120 combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan.

4 0

2 years ago

5 customers buy 10,15,20,25,30 toffees respectively from 1 shop. But the the person who bought 20 toffees forgot to pay .if the

Mice21 [21]

Answer:

40% profit

Step by step Explanation:

Profit percentage

=( profit/cost price) * 100

0.2 = profit/cost

10+15+20+25+30= 100

Let's assume the cost price of the items is $1 each

Cost price total= $100

Profit made when buyer of 20 toffe didn't say was

0.2=profit/cost

0.2*100 =$20

If the$ 20 paid.

Total profit = $40

So percentage profit now

40/100 * 100 = 40%

3 0

2 years ago

Multiply 3xyz ^ 2 / 6y ^ 4 by 2y / x z ^ 4

Marat540 [252]

$\dfrac{3xyz^2}{6y^4}=\dfrac{3}{6}\cdot\dfrac{xyz^2}{y^4}=\dfrac{1}{2}\cdot\dfrac{xz}{y^3}=\dfrac{xz}{2y^3}\\\\\\\dfrac{3xyz^2}{6y^4}\cdot\dfrac{2y}{xz^4}=\dfrac{xz^2}{2y^3}\cdot\dfrac{2y}{xz^4}=\dfrac{2}{2}\cdot\dfrac{xyz^2}{xy^3z^4}=1\cdot\dfrac{1}{y^2z^2}=\dfrac{1}{y^2z^2}$

8 0

3 years ago

Read 2 more answers

Two pounds of bananas cost $1.16 what would be the cost of 5pounds s of bananas

ivann1987 [24]

If 2 lb of bananas cost $1.16, then each pound costs $0.58.

0.58 * 5 = 2.9

So 5 lb of bananas will cost $2.90.

Hope this helps!

6 0

2 years ago

Read 2 more answers