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2 years ago

FREEE POINTSSSSSS HAVE GOOD DAY

SAT

2 answers:

BabaBlast [244]2 years ago

6 0

Answer:

THANKSSS YOU HAVE A GOOD DAY TOOOO

Lelu [443]2 years ago

6 0

thank you sooooooooooooo much :)

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Can someone pls help me with this and if you do all five you get brainiest and more point

lyudmila [28]

Answer:

<h2>How can Carmen improve the Organization and charity of her reason and evidence? </h2>

<h2>Make ADD EXPLANATION </h2>

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2 years ago

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Each month, elizabeth budgets $760 for fixed expenses, $620 for living expenses, and $80 for annual expenses. her annual net inc

kolbaska11 [484]

Answer:

760

Explanation:

it'd the budget that's what it says

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2 years ago

Amos kribbytheking :)

Aleonysh [2.5K]

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Uh sure maybe ?

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2 years ago

The idea that a dollar received in the future has less value than a dollar received today because a dollar received today can ea

Gennadij [26K]

The interest received on a dollar invested over time is called Return on Investment (ROI

<h3>Return on Investment</h3>

It should be noted that if the dollar received today is not put into good investment it might as well be the same as the dollar received in future.

Money losses value overtime hence in other for money to have values it should be invest carefully and wisely.

Learn more about investment here:

brainly.com/question/1305349

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1 year ago

The average waist size for teenage males is 29 inches with a standard deviation of 1. 4 inch. If waist sizes are normally distri

sweet [91]

Using the normal distribution, it is found that 0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

In this problem:

The mean is of $\mu = 29$ .

The standard deviation is of $\sigma = 1.4$ .

The proportion of teenagers who will have waist sizes greater than 31 inches is <u>1 subtracted by the p-value of Z when X = 31</u>, hence:

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

$Z = \frac{31 - 29}{1.4}$

$Z = 1.43$

$Z = 1.43$ has a p-value of 0.9236.

1 - 0.9236 = 0.0764.

0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.

More can be learned about the normal distribution at brainly.com/question/24663213

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2 years ago