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ratelena [41]
4 years ago
6

Jamal ran 2/3 mile. Ming ran 2/4 mile. Liana ran 7/12 of mile. Who ran the farthest? who ran the least?

Mathematics
1 answer:
zmey [24]4 years ago
3 0

Answer:

Jamal ran the farthest.

Ming ran the least.

Step-by-step explanation:

We have to write the number of miles ran by each participant as decimal values, before finding who ran the most and the least.

Important to remember that a fraction is a division, so, for example, 2/3 is 2 divided by 3.

Jamal ran 2/3 mile.

2/3 = 0.667

So Jamal ran 0.667 miles

Ming ran 2/4 mile.

2/4 = 0.5

So Jamal ran 0.5 miles.

Liana ran 7/12 of mile

7/12 = 0.583

So Liana ran 0.583 miles

Who ran the farthest?

Jamal ran the highest amount of miles, so he ran the farthest.

Who ran the least?

Ming ran the least amount of miles, so he ran the least.

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The 3 goes with the square root you don’t need to multiply or anything like that. If you need to simply for example 3 square root of 20 which is when 3 is outside the square root sign. The 20 will be separated with 4*5 in the square root and then you will square root the 4 leaving the 5 in and bringing 2 out. Now you can’t just bring it out and forget about the 3 outside you need to multiply that 2 and 3 and you get 6! So you final answer is 6 square root 5. Hope this helps reply if you have a question!
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John needs to find out the probability that he will sell all his cars by the end of the
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c) 100

Step-by-step explanation:

This is the best choice because the number is not too low or too high. He will get an accurate probability.

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3 years ago
Which of the following inequalities matches the graph? ​
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Answer:

Step-by-step explanation:

y ≤ 5x - 1

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3 years ago
According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month. Assume
Ira Lisetskai [31]

Answer:

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

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.

According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month.

This means that \mu = 22

Standard deviation is 4:

This means that \sigma = 4

Sample of 36:

This means that n = 36, s = \frac{4}{sqrt{36}}

What is the probability that the sample mean is between 21 and 22?

This is the p-value of Z when X = 22 subtracted by the p-value of Z when X = 21.

X = 22

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

By the Central Limit Theorem

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

Z = \frac{22 - 22}{\frac{4}{sqrt{36}}}

Z = 0

Z = 0 has a p-value of 0.5.

X = 21

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

Z = \frac{21 - 22}{\frac{4}{sqrt{36}}}

Z = -1.5

Z = -1.5 has a p-value of 0.0668.

0.5 - 0.0668 = 0.4332

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

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3 years ago
What is the greatest common factor of the polynomial below?
Y_Kistochka [10]

Answer:

Step-by-step explanation:

8x^2 - 4x

4x(2x - 1)

GCF is 4x

5 0
3 years ago
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