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Ratling [72]
3 years ago
10

Attached a picture. Really need some help with this please and thank you

Mathematics
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

c

Step-by-step explanation:

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Please help me with this ​
Gelneren [198K]

345235235232352352353523523535

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3 years ago
What is the value of x?<br> x+10<br> 105
sashaice [31]
X+10=105
105-10=95=x
x = 95
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3 years ago
Write the following equation in standard form: y + 4 = -3(x + 2)
Irina18 [472]

Answer:

-3x-y=10

Step-by-step explanation:

y+4=-3(x+2)

y+4=-3x-6

y=-3x-6-4

y=-3x-10

-3x-y=10

5 0
3 years ago
Pythagorean Theorem Word Problems Mazel​
mixer [17]
Is there supposed to be a picture here?? There isn’t.
5 0
3 years ago
The average Act score follows a normal distribution, with a mean of 21.1 and a standard deviation of 5.1. What is the probabilit
ohaa [14]

Answer:

0.43% probability that the mean IQ score of 50 randomly selected people will be more than 23

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 question:

\mu = 21.1, \sigma = 5.1, n = 50, s = \frac{5.1}{\sqrt{50}} = 0.7212

What is the probability that the mean IQ score of 50 randomly selected people will be more than 23

This is 1 subtracted by the pvalue of Z when X = 23. So

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

By the Central Limit Theorem

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

Z = \frac{23 - 21.1}{0.7212}

Z = 2.63

Z = 2.63 has a pvalue of 0.9957

1 - 0.9957 = 0.0043

0.43% probability that the mean IQ score of 50 randomly selected people will be more than 23

5 0
3 years ago
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