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djverab [1.8K]
3 years ago
5

Solve for y; 2x + 3y = 13 (x = 2)

Mathematics
1 answer:
I am Lyosha [343]3 years ago
4 0

Answer:

y=13x^2/3−2x/3

Step-by-step explanation:

the equal sign in this equation is not meant for this type of problem so this is the answer without the equal sign

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Genrish500 [490]

Answer:

a) 0.0537 = 5.37% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes

b) 0.9871 = 98.71% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes

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.

Mildly obese:

Mean 376 minutes and standard deviation 67 minutes, which means that \mu = 376, \sigma = 67

Sample of 6

This means that n = 6, s = \frac{67}{\sqrt{6}} = 27.35

Lean

Mean 520 minutes and standard deviation 110 minutes, which means that \mu = 520, \sigma = 110

Sample of 6

n = 6, s = \frac{110}{\sqrt{6}} = 44.91

A) What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes?

This is 1 subtracted by the pvalue of Z when X = 420, using the mean and standard deviation for mildly obese people. So

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

By the Central Limit Theorem

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

Z = \frac{420 - 376}{27.35}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463

1 - 0.9463 = 0.0537

0.0537 = 5.37% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes.

B) What is the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes?

This is 1 subtracted by the pvalue of Z when X = 420, using the mean and standard deviation for lean people. So

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

Z = \frac{420 - 520}{44.91}

Z = -2.23

Z = -2.23 has a pvalue of 0.0129

1 - 0.0129 = 0.9871

0.9871 = 98.71% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes

5 0
3 years ago
A number decreased by 7%
Mariulka [41]
X-7/100
OR
X-0.07
Since we don't know the number we need a variable which can be any alphabet so I chose "X".
For example if it were 7 decreased by 7 percent, it would be equal to:
7 - 7/100=6.93
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Answer:

2 and 7 are the prime numbers.

Step-by-step explanation:

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well the answer should be A B to find out how much money they have saved up.

Step-by-step explanation:

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