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r-ruslan [8.4K]
2 years ago
14

Suppose 123 people voted for Danny. If 55 of the votes were from girls, what would be the ratio of votes from girls to votes fro

m boys? Explain.
Mathematics
2 answers:
serious [3.7K]2 years ago
6 0
It would be 55:68, since there are 55 girls to the 68 boys.
Lubov Fominskaja [6]2 years ago
5 0
First, 123 - 55 = 68  which is how many boys voted. So, the ratio is 55:68
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What number should go in the space? Multiplying by 1.98 is the same as increasing by _____%.
sattari [20]
1.98
1.98*100
198

Multiplying by 1.98 is the same as increasing by 198%.
4 0
1 year ago
An apartment complex on Ferenginar with 250 units currently has 223 occupants. The current rent for a unit is 892 slips of Gold-
Triss [41]

Answer:

Step-by-step explanation:

Given data

Total units = 250

Current occupants = 223

Rent per unit = 892 slips of Gold-Pressed latinum

Current rent = 892 x 223 =198,916 slips of Gold-Pressed latinum

After increase in the rent, then the rent function becomes

Let us conside 'y' is increased in amount of rent

Then occupants left will be [223 - y]

Rent = [892 + 2y][223 - y] = R[y]

To maximize rent =

\frac{dR}{dy}=0\\=2(223-y)-(892+2y)=0\\=446-2y-892-2y=0\\=-446-4y=0\\y=\frac{-446}{4}=-111.5

Since 'y' comes in negative, the owner must decrease his rent to maximixe profit.

Since there are only 250 units available;

y=-250+223=-27\\\\maximum \,profit =[892+2(-27)][223+27]\\=838 * 250\\=838\,for\,250\,units

Optimal rent - 838 slips of Gold-Pressed latinum

8 0
2 years ago
Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300.
Fynjy0 [20]

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

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 = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10.

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when X = 500. So

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

Z = \frac{500 - 465}{10}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.

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I hope I've helped
6+1=7?
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