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Arlecino [84]
3 years ago
6

The Sparkle Farm grows both corn and wheat, among other crops. The cost of cultivating an acre of corn and an acre of wheat is $

42 and $30, respectively. The farm has a total of $18,564 in order to cultivate these two crops, however, the farm must grow three times as many acres of corn as wheat in order to qualify forspecific government subsidies. How many acres of wheat must be cultivated
Mathematics
1 answer:
gogolik [260]3 years ago
4 0

Answer: 119

Step-by-step explanation:

Let the acre of wheat to be cultivated be x.

Since the farm must grow three times as many acres of corn as wheat, therefore, corn = 3x

The information given can be used to form a equation as:

42(3x) + 30x = 18564

126x + 30x = 18564

156x = 18564

x = 18564/156

x = 119 acres of wheat

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Answer:

35.00

Step-by-step explanation:

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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

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Which of the following statements is true?<br><br>i need help look at image most appreciated.​
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​At a college, the cost of tuition increased by 10%. Let b represent the former cost of tuition. Use the expression b+0.10b for
kirza4 [7]

Question is Incomplete,Complete question is given below;

At a college, the cost of tuition increased by 10%. Let b be the former cost of tuition. Use the expression b + 0.10b for the new cost of tuition.

a)  Write an equivalent expression by combining like terms.

b)  What does your equivalent expression tell you about how to find the new cost of tuition?

Answer:

a. The equivalent expression is 1.1b.

b. The new cost of tuition is 1.1 times the former cost of tuition.

Step-by-step explanation:

Given:

Former cost of tuition = b

the cost of tuition increased by 10%.

New cost of tuition = b+0.10b

Solving for part a.

we need to find the equivalent expression by combining the like terms we get;

Now Combining the like terms we get;

new cost of tuition = b(1+0.1) = 1.1b

Hence The equivalent expression is 1.1b.

Solving for part b.

we need to to say about equivalent expression about how to find the new cost of tuition.

Solution:

new cost of tuition = 1.1b

So we can say that.

The new cost of tuition is 1.1 times the former cost of tuition.

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