Solve similar triangles (advanced) Solve for x

Mathematics

1 answer:

salantis [7]3 years ago

7 0

Answer:

Step-by-step explanation:

AD=4+8=12=DE thus angle EAD= angle AED=90÷2=45

angle ACB=90-45=45 thus CB=AB=4

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Suppose a jar contains 11 red marbles and

Alina [70]

Answer:

There are 32 marbles. 11 are red. So odds of a red on the first are 11/32.

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Probability both are red, just multiply …. 11/32x 10/31 = 110/992 or 0.111.

8 0

2 years ago

I need help with problem 6) A. Please help.

vampirchik [111]

Answer:

Step-by-step explanation:

The ratios are all 3 father steps to 5 son steps. Then ...

son-steps = 5/3·father steps

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and

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As a <em>check</em>, you can realize that the numbers in the first blank, 36:60, are double those in the second blank, 18:30. The first of these, 36:60, is 3 times the row underneath it, 12:20.

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3 years ago

An elevator has a placard stating that the maximum capacity is 1600 lb�10 passengers. So, 10 adult male passengers can have a m

Bezzdna [24]

Answer:

1. 0.7421 = 74.21% probability the elevator is overloaded.

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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 $s = \frac{\sigma}{\sqrt{n}}$ .