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nlexa [21]
3 years ago
11

What is the value of x/2y, when .x = 10 and y = 17

Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
6 0

Answer:

I think that it might be 5/17

Step-by-step explanation:

2 times 17= 34

10/34= 5/17 or 0.2941176471

REY [17]3 years ago
3 0

Step-by-step explanation:

x=10

y=17

10/2(17)

10/34

ans= 5/17

mark my answer brainliest pliz

Good luck

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Monica deposits $100 into a savings account that pays a simple interest rate of 3.8%. Paul
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Answer:

Paul

Step-by-step explanation:

Paul earn more than monica. Though he is getting less interest but because of his higher initial amount he is getting more return.

Computing the return of both

Monica return is  100*3.4%= $3.4 in a year

Paul return is 200*%2.2=  $4.4 in a year.

7 0
2 years ago
Read 2 more answers
g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib
algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

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

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125

This means that n = 125, s = \frac{7320}{\sqrt{125}}

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{\frac{7320}{\sqrt{125}}}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

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