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7nadin3 [17]
3 years ago
10

How much candy at $1.20 a pound should be mixed with candy worth 95¢ a pound in order to obtain a mixture of 50 pounds of candy

worth a dollar a pound?
Mathematics
1 answer:
Nesterboy [21]3 years ago
5 0

10 pounds of candy costing $1.20 per pounds must be mixed with 40 pounds of candy costing $ 0.95 per pound to obtain a 50 pounds of candy costing $ 1 per pound

<em><u>Solution:</u></em>

Let "x" be the pounds of candy at $ 1.20 per pound

Then (50 - x) is the pounds of candy at $ 0.95 per pound ( since 95 cents is equal to 0.95 pound)

Therefore, x pounds of candy costing $1.20 per pounds must be mixed with (50 - x)  pounds of candy costing $ 0.95 per pound to obtain a 50 pounds of candy costing $ 1 per pound

Then the equation becomes,

1.20x + (50 - x)0.95 = 50 x 1

On expanding we get,

1.20x + 47.5 - 0.95x = 50

0.25x = 50 - 47.5

0.25x = 2.5

x = 10

Then (50 - x) = 50 - 10 = 40

So we need 10 pounds of candy worth $ 1.20 per pound and 40 pounds of candy worth $ 0.95 to obtain 50 pounds of candy worth a dollar a pound

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Answer:

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

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}}

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The randomly sampled 400 dogs from homes where an herbicide was used on a regular basis, diagnosing lymphoma in 230 of them.

This means that:

p_h = \frac{230}{400} = 0.575, s_h = \sqrt{\frac{0.575*0.425}{400}} = 0.0247

Of 200 dogs randomly sampled from homes where no herbicides were used, only 25 were found to have lymphoma.

This means that:

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Distribution of the difference:

p = p_h - p_n = 0.575 - 0.125 = 0.45

s = \sqrt{s_h^2+s_n^2} = \sqrt{0.0247^2 + 0.0234^2} = 0.034

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The confidence interval is:

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So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

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The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

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