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3 years ago

Find A intersection B.

1 answer:

3 0

A: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
B: {-2, -1, 1, 4, 7, 11}

A ∩ B = {1, 4, 7, 11}

So the answer is B.

Intersection of two or more sets cointains only common elements.

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A gift box has the shape of a cube with an edge length of 4.5 in.

BlackZzzverrR [31]

91.125 all you need to do is cube 4.5 or multiply 4.5 by itself 3 times so 4.5x4.5x4.5

8 0

3 years ago

Read 2 more answers

Sarah uses a total of 36 sheets of paper for her story. Each chapter is 9 pages , how many pages are in her short story

MariettaO [177]

Answer:

4 pages

Step-by-step explanation:

You divide 36 by 9

4 0

2 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So $n = 6, s = \frac{106}{\sqrt{6}} = 43.27$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 523}{43.27}$

$Z = -2.61$

$Z = -2.61$ has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0

3 years ago

A = 4 and b = 2 whoever answers first gets brainliest and 50 points

ankoles [38]

Answer:

Step-by-step explanation:

6 0

3 years ago

A daily mail is delivered to your house between 1:00 p.m. and 5:00 p.m. Assume delivery times follow the continuous uniform dist

german

Answer:

25% of deliveries

Step-by-step explanation:

This is a uniform distribution with parameters a = 1.00 and b =5.00. With these conditions, the following probability distribution functions can be applied:

$P(X \leq x)=\frac{x-a}{b-a}=\frac{x-1.00}{5.00-1.00}\\P(X \leq x) = \frac{x-1}{4}\\P(X \geq x) = 1- P(X \leq x) \\P(X \geq x)=1- \frac{x-1}{4}$

Therefore, the probability that X ≥ 4:00 p.m. is given by

$P(X \geq 4)=1- \frac{4-1}{4}\\P(X \geq 4)=0.25=25\%$

Thus, 25% of deliveries are made after 4:00 p/m.

7 0

3 years ago