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tangare [24]
3 years ago
15

The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items

are produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, C(x)?
Mathematics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

a. $500

b. $250

Step-by-step explanation:

a. For zero items produced, cost will be-

C(0) = 10(0) + 500 = 0 + 500 = 500

b. For zero items produced, cost will be-

C(25) = 10(25) + 500 = 250 + 500 = 750

c. If C(x)_{max} = 1500

10x + 500 = 1500\\10x = 1000\\x = 100

∴ domain for cost function will be [0,100] and range will be 100

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First you need to make both bases the same:

Lets remove the ^p and ^4

To make the base of 42 equal to 41, you would have 41^x = 41

X - ln(42) / ln(41) = 1.00648904


Now you have 41^1.00648904(p) = 41^4


Now the bases are equal so we need to set the exponents to equal:

1.00648904(p) = 4

Divide both sides by 1.00648904 to solve for P

P = 4 / 1.00648904

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Answer:

At 95% confidence level, the difference between the proportion of freshmen using steroids in Illinois and the proportion of seniors using steroids in Illinois is -7.01135×10⁻³ < \hat{p}_1-\hat{p}_2 < 1.237

Step-by-step explanation:

Here we are required to construct the 95% confidence interval of the difference between two proportions

The formula for the confidence interval of the difference between two proportions is as follows;

\hat{p}_1-\hat{p}_2\pm z^{*}\sqrt{\frac{\hat{p}_1\left (1-\hat{p}_1  \right )}{n_{1}}+\frac{\hat{p}_2\left (1-\hat{p}_2  \right )}{n_{2}}}

Where:

\hat{p}_1  = \frac{34}{1679}

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Plugging in the values, we have;

\frac{34}{1679}- \frac{24}{1366} \pm 1.96 \times \sqrt{\frac{ \frac{34}{1679}\left (1- \frac{34}{1679}\right )}{1679}+\frac{\frac{24}{1366} \left (1-\frac{24}{1366}   \right )}{1366}}

Which gives;

-7.01135×10⁻³ < \hat{p}_1-\hat{p}_2 < 1.237.

At 95% confidence level, the difference between the proportion of freshmen using steroids in Illinois and the proportion of seniors using steroids in Illinois = -7.01135×10⁻³ < \hat{p}_1-\hat{p}_2 < 1.237.

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