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Kryger [21]
3 years ago
9

Which statement about g(x)=x^2-576 is true

Mathematics
1 answer:
goldenfox [79]3 years ago
5 0
We can answer without the picture or answer options
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SOMEONE HELP ME PLEASE
marin [14]

Answer:

Step-by-step explanation:

Direct variation problems can easily be solved with proportions, namely:

\frac{x}{y} :\frac{2}{5}=\frac{4}{y} and cross multiply to get

2y = 20 so

y = 10

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The length of a book cover are 22.2 centemeters respectively
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Madeline bought a new coat at a 20% off sale. She save $27. What was the original price of the coat?
kompoz [17]

Let, the original price = x

x*20 /100 = 27

x = 2700/20 = 135

Original price was $135
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3 years ago
If you<br> earn 96 aweek<br> that<br> How much<br> month
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Answer:

384

Step-by-step explanation:

since there are like 4 weeks in a month  so u do

96 multiplied by 4 =384

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3 years ago
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Suppose 44% of the children in a school are girls. If a sample of 727 children is selected, what is the probability that the sam
Harman [31]

Answer:

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Suppose 44% of the children in a school are girls.

This means that p = 0.44

Sample of 727 children

This means that n = 727

Mean and standard deviation:

\mu = p = 0.44

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.44*0.56}{727}} = 0.0184

What is the probability that the sample proportion of girls will be greater than 41%?

This is 1 subtracted by the p-value of Z when X = 0.41. So

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

By the Central Limit Theorem

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

Z = \frac{0.41 - 0.44}{0.0184}

Z = -1.63

Z = -1.63 has a p-value of 0.0516

1 - 0.0516 = 0.9884

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

5 0
3 years ago
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