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Mariulka [41]
3 years ago
12

What is the measure of the indicated angle? 120° ?

Mathematics
1 answer:
aliya0001 [1]3 years ago
8 0
I think the answer is 180
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What is the inverse of f(x)=x^2 - 2
Allushta [10]

Answer:

f^{-1} (x) = \sqrt{x+2} , - \sqrt{x+2}

Step-by-step explanation:

To find the inverse, interchange the variables and solve for y.

Hope this helps,

Davinia.

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The coach bought a hamburger and a $1.75 drink for the 10 members of the team. He spent a total of $36.00. Write and nu solve an
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1.85

Step-by-step explanation:

equation will be 10(h+1.75)=36

hamburger price is $1.85

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What is number 315,962,004 write in words ​
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three hundred  fifteen million nine hundred sixty two thousand four

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An electrician charges a $100 flat fee, plus $75 per hour to work on a house.
jek_recluse [69]

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Step-by-step explanation:

8 0
2 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
3 years ago
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