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Salsk061 [2.6K]
3 years ago
11

The Montanez family is a family of four people. They have used 3,485.78 gallons of water so far this month. They cannot exceed 7

,250.50 gallons per month during drought season. Write an inequality to show how much water just one member of the family can use for the remainder of the month, assuming each family member uses the same amount of water every month.
Mathematics
1 answer:
faust18 [17]3 years ago
8 0

Answer: yes

Step-by-step explanation:

this month 3485.78/4 = 871.445 gallons per person

Drought month 7250.50/4 = 1812.625

difference 7250.50 - 3485.78 = 3764.72

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What is 2/3 Time 2/3
klasskru [66]

Answer:

\frac{4}{9}  

Step-by-step explanation:

\frac{2}{3} * \frac{2}{3}

you multiply across, so you do 2*2 for the numerator, then 3*3 for the denominator. This would equal 4 as the numerator, 9 as the denominator,  which is \frac{4}{9}  

4 0
3 years ago
Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample
fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 estabilishes 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}}

37% of the company's orders come from first-time customers.

This means that p = 0.37

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

\mu = p = 0.37

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

Z = \frac{0.38 - 0.37}{0.0322}

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 0.26

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

Z = \frac{0.26 - 0.37}{0.0322}

Z = -3.42

Z = -3.42 has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

5 0
2 years ago
A shipping container will be used to transport several 70-kilogram crates across the country by rail. The greatest weight that c
alina1380 [7]

Subtract what has already been loaded from total amount allowed:

23500 - 9700 = 13,800

Divide the amount left by the weight of each crate:

13,800 / 70 = 197.14

They can load 197 crates.

6 0
3 years ago
Algebra 1!! Please only answer if you actually know!
natka813 [3]

Answer:

9. B

y=-5 and x=-25

y=(1/5)x

-5=(1/5)*-25

-5=-5

10. A

y=42 and x=14

y=3x

42=36*14

42=42

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A bank offers loans with simple interest rates for those who qualify. If the principal is the same for each, which two plans cha
Julli [10]
20is going to be the answer
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