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3 years ago

A crew is picking up chairs in the gymnasium after an event. After clearing 68 chairs from

Mathematics

2 answers:

solong [7]3 years ago

8 0

Answer:

Foyjffjfjfjfjfkffkfjfjjffjjfkfjf 8rruirkteiuiffk

saw5 [17]3 years ago

4 0

Answer:

68+14=c c=82

Step-by-step explanation:

just add 68 and 14 to get the answer 82

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Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems o

NISA [10]

Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so $\pi = 0.8$ .

(a) What is the probability she gets an A?

There are four problems, so $n = 4$

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find $P(X \geq 3)$

$P(X \geq 3) = P(X = 3) + P(X = 4)$

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

$P(X = 3) = C_{4,3}.(0.80)^{3}.(0.2)^{1} = 0.4096$

$P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192$

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so $n = 3$

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find $P(X \geq 2)$

$P(X \geq 3) = P(X = 2) + P(X = 3)$

$P(X = 2) = C_{3,2}.(0.80)^{2}.(0.2)^{1} = 0.384$

$P(X = 4) = C_{3,3}.(0.80)^{3}.(0.2)^{0} = 0.512$

$P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896$

If she gets the first problem correct, there is an 89.6% probability that she gets an A.

3 0

3 years ago

Ayudenme 5fjgfgjzngfjgxj kcnhgbxjhdjtfbgrbftjxbxh

olchik [2.2K]

Ayudenme 5fjgfgjzngfjgxj

kcnhgbxjhdjtfbgrbftjxbxh

6 0

3 years ago

(5, 3) is the solution to the system of linear equations: True False

Mkey [24]

I believe it’s true!

5 0

3 years ago

Read 2 more answers

Phones-R-Us charges $16.95 per month and $0.05 per text message. Awesome Wireless charges $22.95 per

Dima020 [189]

Answer:

$S \leq 200$

Step-by-step explanation:

Given

Phone R-Us= $16.95 + $0.05 per SMS

Awesome Wireless = $22.95 + $0.02 per SMS

Required

Determine the number of SMS such that Awesome Wireless is greater or equal to Phone R-Us

Represent the SMS with S

For Phone R-Us, we have:

$SMS = 16.95 + 0.05S$

For Awesome Wireless, we have:

$SMS = 22.95 + 0.02S$

For Awesome Wireless is greater or equal to Phone R-Us, we have:

$22.95 + 0.02S \geq 16.95 + 0.05S$

Collect Like Terms

$0.02S - 0.05S \geq 16.95 - 22.95$

$-0.03S \geq -6$

Solve for S

$\frac{-0.03S}{-0.03} \geq \frac{-6}{-0.03}$

$S \leq 200$

Hence: for Awesome Wireless to cost more or equal to Phone R-Us, the number of SMS must not exceed 200

4 0

3 years ago

Can you please answer these 3 questions im having a hard time rn and i just need to relax but i really need to get this done, pl

Yanka [14]

Answer:

1. Infinitiely many solutions, one solution, no solutions

y=-3/2x-4

y= x+1

y=1/4x-2

y=1/4x+3

Step-by-step explanation:

4 0

3 years ago