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Nat2105 [25]
2 years ago
5

PLEASE HELP!! I NEED TO PASS MATH

Mathematics
1 answer:
BARSIC [14]2 years ago
6 0

Answer:

A pair of vertical angles would be 9 and 7

Step-by-step explanation:

a relationship between them would be that they equal the same amount.

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Stephanie has 6.98 lb of granola to divide into small bags. She will put 0.4 lb of granola in each bag.
Vsevolod [243]
When you divide 6.98 lbs. and 0.4 lbs, you will get the answer of 17.45
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3 years ago
Suppose f(x)=x+3 find the graph of f(2x)
dezoksy [38]

Answer:

<h2>2x + 3</h2>

Step-by-step explanation:

f(2x) means that we have to substitute x with 2x

So, the answer is

2x + 3

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2 years ago
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Answer:

ok

Step-by-step explanation:

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3 0
2 years ago
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The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
Y=5x+2 and 3x=-y+10 what is the soluion of equation
Harlamova29_29 [7]
I hope this is the answer you want

4 0
3 years ago
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