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Fiesta28 [93]
3 years ago
7

Wich pair of equations below represents perpendicular lines? Please!!!

Mathematics
2 answers:
bulgar [2K]3 years ago
7 0

B) y = 5x + 15 and y = -5x + 15 I think.

sp2606 [1]3 years ago
3 0

Answer:

It is A

Step-by-step explanation:

it is the negative recipricol

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Tyler believes that an 8-sided die can be used to predict whether customers at his store will use a coupon when making their pur
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"1 indicating a coupon and all other outcomes indicating no coupon"
Probability is (number of successful outcomes) / (number of possible outcomes)

Theoretical Probability of rolling a 1: 1/8 

Experimental Probability of using coupons: 4/48  =  1/12

So, the experimental probability of a customer using a coupon (that is, 1/12) is smaller than the theoretical probability of rolling a 1 (that is, 1/8).
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Solve:
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a= -5

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At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine
Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

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 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.

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}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

Mean and standard deviation:

\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

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

By the Central Limit Theorem

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

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

8 0
2 years ago
I NEED HELP FAST!!!!!!!!!! ⦁ Solve the equation. Describe the solution to the equation as one solution, no solution, or an infin
g100num [7]
\frac{1}{8} (x + 24) =  \frac{5}{8} x + 9 -  \frac{1}{2} x \\ \\  \frac{x + 24}{8} =  \frac{5}{8} x + 9 -  \frac{1}{2} x \ / \ simplify \\ \\  \frac{x + 24}{8} =  \frac{5x}{8} + 9 - \frac{1}{2} x \ / \ simplify \\ \\  \frac{x+ 24}{8} =  \frac{5x}{8} + 9 -  \frac{x}{2} \ / \ simplify \\ \\  \frac{x + 24}{8} =  \frac{x}{8} + 9 \ / \ simplify \\ \\ x + 24 = x + 72 \ / \ multiply \ each \ side \ by \ 8 \\ \\ 24 = 72 \ / \ cancel \ x \\ \\ no \ solution \\ \\

The final result is: no solution.
7 0
3 years ago
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