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4 years ago

An equation is shown. 2x-4y=8 solve the equation for x in terms of y

1 answer:

7 0

2x-4y=8

Add 4y to each side to separate the variables, since you want to solve for x let be on the left

2x=4y+8

Divide the entire equation by 2 to allow x to stand alone

x=2y+4

And I have to type some more so it will let me answer the question so here ya go

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PLS HELP ASAP WILL MARK BRAINLIEST Determine whether each relation is a function. Explain your reasoning.

Mariana [72]

No, it is not a function, since 6 is repeating itself and the x should never repeat if it’s a function.

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3 years ago

Logan and three friends went out to dinner. The total bill came to $74.40. They decide to split the bill equally between them. H

olya-2409 [2.1K]

74.40/3 friends= 24.80 for each friend. Simply division, easy to do on a calculator if you are allowed one.

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4 years ago

Read 2 more answers

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$

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.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when $X = 59.5$ . So

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

$Z = \frac{59.5 - 56}{1.12}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0

3 years ago

A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Ke

yaroslaw [1]

Answer:

a. 0.563 = 56.3% probability that for the next 60 minutes (two time periods) the system will be in the delay state.

b. 0.625 = 62.5% probability that in the long run the traffic will not be in the delay state

Step-by-step explanation:

Question a:

The probability of finding a traffic delay in one period, given a delay in the preceding period, is 0.75.

The system currently is in traffic delay, so for the next time period, 0.75 probability of a traffic delay. If the next period is in a traffic delay, the following period will also have a 0.75 probability of a traffic delay. So

0.75*0.75 = 0.563

0.563 = 56.3% probability that for the next 60 minutes (two time periods) the system will be in the delay state.

b. What is the probability that in the long run the traffic will not be in the delay state? If required, round your answers to three decimal places.

If it doesn't have a delay, 85% probability of continuing without a delay.

If it has a delay, 75% probability of continuing with a delay.

So, for the long run:

x: current state

85% probability of no delay if x is in no delay, 100 - 75 = 25% if x is in delay(1-x). So

$0.85x + 0.25(1 - x) = x$