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ZanzabumX [31]
3 years ago
11

I don't get this at all help me please

Mathematics
2 answers:
hodyreva [135]3 years ago
8 0
2937728488 yes3784 666
Vikentia [17]3 years ago
5 0

Answer:

The answer is correct

Step-by-step explanation:
Disclaimer: x will represent b

The reason that it is correct is because it says that "each also buys a healthy snack bag". If we look at the problem, we know that each friend needed a ticket to get in, which cost 6.75. Let's say that the price of the Healthy Snack Bag is x. We know that four friends go together, and each has to pay a ticket. So we have

4 Friends each get a healthy snack bag with the price of x and a ticket that costs 6.75.

So the price that each friend had to pay was 6.75(ticket) + x(snack)
All four friends needed to do that, so we have 4 of the top equation. That can simplify to 4(6.75+x). It also says that the total cost of all 4 of them was 37. So 4(6.75+x) = 37. By using the distributive property, you get 27+4x=37.
With some simple algebra, and by isolating x, you get x = 2.5.

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Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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}}

Proportion of 0.6

This means that p = 0.6

Sample of 46

This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.6}{0.0722}

Z = -1.38

Z = -1.38 has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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