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marusya05 [52]
3 years ago
5

Helpppppppppppppppppp Pppppo

Mathematics
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

(2 * L)+(2 * W)

Step-by-step explanation:

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Sorry, I don’t know lol

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the vega family has a cell phone plan that costs $75 per month including taxes and fees. the plan lets the 5 members of the vega
torisob [31]
Sent a picture of the solution to the problem (s).

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3 years ago
Simplify the expression-2.3 + (-5.7). On the test when maureen simplified the expression she got -3.4. What mistake did maureen
Hoochie [10]

Answer:

see explanation

Step-by-step explanation:

- 2.3 + (- 5.7)

reminder that + (- ) = -

To obtain - 3.4 it is likely that she added 2.3 to - 5.7

The solution is

- 2.3 - 5.7 = - 8

4 0
3 years ago
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The mean height of women in a country (ages 20-29) is 64 4 inches A random sample of 50 women in this age group is selected What
Simora [160]

Answer:

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 64.4 inches, standard deviation of 2.91

This means that \mu = 64.4, \sigma = 2.91

Sample of 50 women

This means that n = 50, s = \frac{2.91}{\sqrt{50}}

What is the probability that the mean height for the sample is greater than 65 inches?

This is 1 subtracted by the p-value of Z when X = 65. So

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

By the Central Limit Theorem

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

Z = \frac{65 - 64.4}{\frac{2.91}{\sqrt{50}}}

Z = 1.46

Z = 1.46 has a p-value of 0.9279

1 - 0.9279 = 0.0721

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

8 0
3 years ago
Which of the following situations can be modeled by the expression -4 + (-7)?
cupoosta [38]

Third option: Elsa lost four points on the last math quiz, and Marc lost seven points. How many total points did they lose?

5 0
3 years ago
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