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Paha777 [63]
3 years ago
11

30 Points! please help!

Mathematics
1 answer:
SVEN [57.7K]3 years ago
3 0

Answer:

18) 4 * 10 ^ -10

19) t ^ 9

Step-by-step explanation:

18. When multiplying in scientific notation, just multiply the regular numbers together and add the exponents for the tens.

So it'll be 40 * 10^-11

However, the beginning number should have the decimal right after the 4, not any place after (for example, it's 3.056, not 30.56 or 3056.)

To fix this, move the decimal place forward one space from 40. to 4.0, and add 1 to the -11 power.

So the answer is now 4 * 10^-10

19. When dividing exponents, if the base number is the same, you can divide. (Basically, you can not divide x^2 and y^3 because x and y aren't the same)

Since in this case it's t, you can divide.

Dividing exponents is simple: just subtract them.

14 - 5 = 9

so the answer is t ^ 9

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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.

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

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