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PilotLPTM [1.2K]
2 years ago
10

A total of d dollars was donated to 4 charities. Each charity received $375. Which equation can be solved to find the total amou

nt of money donated?
Mathematics
1 answer:
lidiya [134]2 years ago
6 0

Answer:

If each charity has received $375 and there is 4 charities, the equation should look like this...

$375 x 4 = d

If you are looking how much was donated all together, the answer is:

$1500 dollars in total was donated.

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Solve for r.

You want to get r by itself on one side on the equal sign.

bh + hr = 25

Subtract bh from both sides.

hr = 25 - bh

Divide h on both sides.

r = 25 - bh / h

The two h's cancel each other out.

r = 25 - b

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When 54,702 is divided by 25, the quotient is 2,188 R ___.
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Hi. Please I need help with this :
ki77a [65]

Answer:

1.79553

Step-by-step explanation:

Step 1: Write expression

log₁₀(√(69.5² - 30.5²))

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2 years ago
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REY [17]

Answer:

0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Distribution of years of experience:

Mean 3, so \mu = 3

Standard deviation 2, so \sigma = 2

Sample of 4:

n = 4, s = \frac{2}{\sqrt{4}} = 1

What is the probability that the mean years of experience from the sample of 4 is greater than 3.5?

1 subtracted by the pvalue of Z when X = 3.5. So

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

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

Z = \frac{3.5 - 3}{1}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.

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2 years ago
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