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Pie
3 years ago
6

I really need help with this question so can someone really help me wit it and break it down to me so I could understand

Mathematics
1 answer:
Ber [7]3 years ago
5 0

Step-by-step explanation:

First, there are two pounds and each pound is 85 cents. So, multiple 85 cents by 2.

That, will come put to $1.70

Now we still have .75 left of a pound. It will be best to figure out how much a quarter pound will cost because 3 of how much one quarter pound cost is .75 pounds. So, in order to find that out we know .85 cents is one pound so we will have to dived that by 4.

That will come out to, .21 cents

Now, that is for one quarter pound. we have to find .75 pounds. So, multiple the quarter pound by 3.

That will come out to, .63 cents.

.63 is for .75 of a pound. So, the last thing we have to do is take our .75 pound and add it to our 2 pound one that we solved for earlier.

That will come out to, $2.33

So you're answer will be $2.33 for 2.75 pounds.

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According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0
2 years ago
Find X in this picture. ​
sweet [91]

Answer:

x is in the top left of the triangle :)

Step-by-step explanation:

7 0
3 years ago
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Please solve 1 through 6
Degger [83]
1. 3x³-2x²+1x
2. 6y⁴+2y³-8y
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7 0
3 years ago
HELP PLEASE
daser333 [38]

Answer:

I think its C. I'm not sure I havent been thought this lesson.

7 0
3 years ago
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puteri [66]

Answer:

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Step-by-step explanation:

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