20 POINTS AND BRAINLIEST!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

wolverine [178]3 years ago

7 0

Look for points with whole number coefficients. (2,4) is such a point; (0,1) is another. To find points on the graph of the inverse of this function, all we have to do in this case is to reverse the order of the coefficients, so

(2,4) becomes (4,2) and (0,1) becomes (1,0). Only (4,2) is included in the list of answer choices, above, so the correct answer is (4,2).

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ZA and ZB are vertical angles. If mZA= (5x – 4) and mZB = (2x – 1)º, then find the value of x. INI

natulia [17]

Answer:

x = 1

Step-by-step explanation:

It is given that,

∠A and ∠B are vertical angles.

∠A = (5x – 4) and ∠B = (2x – 1)

We need to find the value of x.

We know that, the vertical angles are equal. So,

∠A=∠B

(5x – 4)=(2x – 1)

Taking like terms together,

5x-2x = 4-1

3x=3

x = 1

So, the value of x is equal to 1.

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3 years ago

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stepan [7]

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natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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