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Maksim231197 [3]
3 years ago
7

Mindy is 39 years old. Mindy's age is 3 years older than two times Jake's age. How old is Jake?

Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
8 0

Answer:

Jake's age = 18.

Step-by-step explanation:

Given  : Mindy is 39 years old. Mindy's age is 3 years older than two times Jake's age.

To find : How old is Jake.

Solution : We have given Mindy is 39 years old.

Let the Jake's age = x .

According to question :

Mindy's age is 3 years older than two times Jake's age.

Mindy's age = 2x  ( Two times of Jake's age )

Mindy's age = 2x + 3 (  Three more than Jake's age ) .

39 = 2x + 3 .

On subtracting both sides by 3

39 -3 = 2x .

36 = 2x .

On Dividing both sides by 2 .

18 = x

Therefore, Jake's age = 18.

Rzqust [24]3 years ago
4 0
The answer would simply be 1
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Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

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.

Mean of 15.4 inches, and standard deviation of 3.5 inches.

This means that \mu = 15.4, \sigma = 3.5

16 items are chosen at random

This means that n = 16, s = \frac{3.5}{\sqrt{16}} = 0.875

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

By the Central Limit Theorem

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

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0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

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olga nikolaevna [1]

ANSWER

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EXPLANATION

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