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3 years ago

Please help asap! (:

Mathematics

2 answers:

tiny-mole [99]3 years ago

8 0

Hello there! Let's calculate this. :)

8u^3+8u^2+6+4u^3−6u+3

=8u^3+8u^2+6+4u^3+−6u+3

Combine Like Terms:

=8u^3+8u^2+6+4u^3+−6u+3

=(8u^3+4u^3)+(8u^2)+(−6u)+(6+3)

=12u^3+8u^2+−6u+9

Answer:

= D. 12u^3+8u^2−6u+9

Cheerio and have a lovely day! ^.^

MAXImum [283]3 years ago

6 0

Hello!

You can only add variables with the same base and exponent

8u^3 + 4u^3 = 12u^3
8u^2 + 0 = 8u^2
0 + -6u = -6u
6 + 3 = 9

Put the sums together

12u^3 + 8u^2 - 6u + 9

The answer is $D) 12u^3 + 8u^2 - 6u + 9$

Hope this helps!

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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

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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.

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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

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