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Ierofanga [76]
3 years ago
7

When renting a limo for prom, the number of people varies inversely with the cost per person. Originally there were 12 people an

d the cost per person was $20. If the number of people changed to 6, what would be the new cost per person?
Mathematics
1 answer:
pashok25 [27]3 years ago
8 0
The answer would be $10.
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What’s the expression of this in simplest form
Ad libitum [116K]

Answer:

3/17

Step-by-step explanation:

3/4 divided by 4  1/4 is = to 3/4 divided by 17/4= 3/4 times 4/17 = 3/17

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(-1/4x-2/3)-(1/2x+1/6) show work no troll answers mark brainlest.
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Answer:

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Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce
LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 ​kg.

92.82% of the sample means are between 75 kg and 83 ​kg.

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}}.

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:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that \mu = 79, \sigma = 22.

What percentage of individual adult females have weights between 75 kg and 83 ​kg?

This percentage is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 75. So:

X = 83

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

Z = \frac{83 - 79}{22}

Z = 0.18

Z = 0.18 has a pvalue of 0.5714.

X = 75

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

Z = \frac{75- 79}{22}

Z = -0.18

Z = -0.18 has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 ​kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each​ sample, what percentage of the sample means are between 75 kg and 83 ​kg?

Now we use the Central Limit THeorem, when n = 100. So s = \frac{22}{\sqrt{100}} = 2.2.

X = 83

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

Z = \frac{83 - 79}{2.2}

Z = 1.8

Z = 1.8 has a pvalue of 0.9641.

X = 75

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

Z = \frac{75-79}{2.2}

Z = -1.8

Z = -1.8 has a pvalue of 0.0359.

This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 ​kg.

8 0
3 years ago
larry wants to buy some carpeting for his living room. the length of the room is 2 times the width and the total area of the roo
gayaneshka [121]
So here is the solution of the given problem above.
To find the area we are just going to use this formula. A = L x W.
Let x be the width and 2x is the length. Given that the area is 8 square meters. So let us find x first.
Plug in the values:
8 = (2x)(x)
8 = 2x^2 << divide both sides by 2 and we get
4 = x^2
Then square 4 and x^ 2 and we got
2 = x
Therefore, the width of the room is 2 meters and the length 2x is 4 meters. Hope this is the answer that you are looking for.

4 0
3 years ago
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