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zhuklara [117]
3 years ago
10

PLEASE HELP ME!!! THIS IS TIMED!!!! PLEASE HELP ME!!!!!!!!!!

Mathematics
2 answers:
wlad13 [49]3 years ago
8 0
I’m almost certain it is 2%
allsm [11]3 years ago
7 0

Answer:

2%

Step-by-step explanation:

if its wrong im so sorry

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Ivan
When x=1 y=13 when x=3 y=19 and when x=5 y=25 you just plug the numbers in for x and solve
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2 years ago
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Suppose the heights of the members of a population follow a normal distribution. if the mean height of the population is 65 inch
Ulleksa [173]

Answer:

D) 56 inches to 74 inches

Step-by-step explanation:

By the Empirical Rule, 99.7% of data in a normal distribution spread across ±3σ standard deviations from the mean μ.

Hence, the range maximum is μ+3σ = 65+3(3) = 65+9 = 74 inches, and the range minimum is μ-3σ = 65-3(3) = 65-9 = 56 inches

Thus, 99.7% of the population will have a height within the range of 56 inches to 74 inches

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2 years ago
Choice test you know the answer to the third question is not b or c you sure about a,d,or e. If you guess, what is the probabili
MrRissso [65]

If there were just one question, then the probability of guessing correctly would be 1/3.

Since all the answers are independent (the answer to one question has no bearing on the answers to the others), then this is the case with each question, so the chances of guessing all answers correctly is 1/3 × 1/3 × 1/3 = 1/27. Independent choices are linked by multiplication.

To have exactly 2 answers correct, we have to think of which one is wrong: there are 3 questions and any single one could be wrong. The probability that the first question is wrong is 2/3. And we know that the probabilities of the other two being right is 1/3 each, so the probability of just the first question being wrong and the others right is 2/3 × 1/3 × 1/3 = 2/27. But this is just one of the three cases: 1/3 × 2/3 × 1/3 and 1/3 × 1/3 × 2/3 also both equal 2/27 each. So here we add the cases together: 2/27 + 2/27 + 2/27 = 6/27 = 2/9. So the answer to part (a) of your question is 2/9.

To solve (b) Consider that (b) is the same as (a) with "all answers right" added in. So you can simply add answer (a) to the chance of guessing all answers correctly.

To solve (c) it might help you to think of the equivalent problem: what is the probability of getting 0 correct plus the probability of getting 1 correct?

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Where is the table and questions Sherlock
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The first and third :)
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