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2 years ago

7

Point I is on line segment HJ. Given I J equals 3X +3, HI equals 3X-1, and HJ equals 3X +8, determine the numerical length of HJ

.

2 answers:

Nezavi [6.7K]2 years ago

4 0

Answer:

Step-by-step explanation:

3x + 3 + 3x - 1 = 3x + 8

6x + 2 = 3x + 8

3x + 2 = 8

3x = 6

x = 2

HJ= 3(2) + 8 =6 + 8 = 14

HI= 3(2) - 1 = 6 - 1 = 5

IJ = 3(2) + 3 = 6 + 3 = 9

Salsk061 [2.6K]2 years ago

4 0

Answer:14

Step-by-step explanation:

3x + 3 + 3x - 1 = 3x + 8

6x + 2 = 3x + 8

3x + 2 = 8

3x = 6

x = 2

HJ= 3(2) + 8 =6 + 8 = 14

HI= 3(2) - 1 = 6 - 1 = 5

IJ = 3(2) + 3 = 6 + 3 = 9

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3 years ago

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3 years ago

How do i do this question? someone help

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8 0

3 years ago

Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr

bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use 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 109.0 inches, and a standard deviation of 12 inches.

This means that $\mu = 109, \sigma = 12$

Sample of 25.

This means that $n = 25, s = \frac{12}{\sqrt{25}} = 2.4$

What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches?

This is the p-value of Z when X = 112. So

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

By the Central Limit Theorem

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

$Z = \frac{112 - 109}{2.4}$

$Z = 1.25$

$Z = 1.25$ has a p-value of 0.8944.

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

7 0

3 years ago