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love history [14]
2 years ago
14

Find the value of x, given that the two triangles are similar.​

Mathematics
2 answers:
8_murik_8 [283]2 years ago
7 0

Answer:

10.5 is the correct answer.

Step-by-step explanation:

Due to the fact Triangle CAB has "6" as its length, and Triangle FDE has "2" as its length, you multiply 3.5 and 3.

3.5 x 3=10.5

2 x 3=6

3.5/2=10.5/6, so the answer is 10.5.

Vlada [557]2 years ago
3 0

Answer:

<h2><u><em>x = 10.5 cm</em></u></h2>

Step-by-step explanation:

Find the value of x, given that the two triangles are similar.

​

Given that the two triangles are similar we can solve with a proportion

6 : 2 = x : 3.5

x = 6 * 3.5 : 2

x = 21 : 2

x = 10.5

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6. Factor 4x from 22x2 + 40x 6.​

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The mean number of words per minute (WPM) read by sixth graders is 97 with a standard deviation of 19 WPM. If 75 sixth graders a
andrey2020 [161]

Answer:

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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 normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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.

In this problem, we have that:

\mu = 97, \sigma = 19, n = 75, s = \frac{19}{\sqrt{75}} = 2.1939

What is the probability that the sample mean would be greater than 101.63 WPM?

This is 1 subtracted by the pvalue of Z when X = 101.63. So

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

By the Central Limit Theorem

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

Z = \frac{101.63 - 97}{2.1939}

Z = 2.11

Z = 2.11 has a pvalue of 0.9826.

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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Maria charges $15 for 3 h of babysitting. How much does Maria charge for 5 h of babysitting?
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