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storchak [24]
3 years ago
6

Budget planners in two towns, Alphaville and Betaville, developed models to determine the budget surplus (in dollars) for a year

based on the tax revenue (in thousands of dollars) for the year. Using historical data, Alphaville’s planner produced the model -2x2+ 500x, while Betaville’s planner produced the model x2- 100x + 10,000. What expression gives how much greater Alphaville’s annual budget surplus is than Betaville’s for a particular amount of tax revenue? If the tax revenue that year in each town is $75,000, how much greater is Alphaville’s budget surplus than Betaville’s that year?
Please help! I WILL MARK BRAINLIEST
Mathematics
1 answer:
kramer3 years ago
3 0

Answer:

(a)-3x^2+150x-10000

(b)-\$16,863,760,000

Step-by-step explanation:

Alphaville's Budget Surplus Model is -2x^2+ 500x

Betaville's Budget Surplus Model is x^2- 100x + 10000.

We want to determine the expression that shows how much greater Alphaville’s annual budget surplus is than Betaville’s for a particular amount of tax revenue.

  • To do this, we subtract Betaville's Model from Alphaville's model.

-2x^2+ 500x-(x^2- 100x + 10000)\\

Opening the brackets

-2x^2+ 500x-x^2+100x - 10000\\

Collect like terms and simplify

-2x^2-x^2+100x + 500x- 10000\\=-3x^2+150x-10000

The expression that shows how much greater Alphaville's Budget is:

-3x^2+150x-10000

(b) If the tax revenue that year in each town is $75,000

We want to evaluate the expression derived above when the tax revenue that year in each town is $75,000 i.e.at x=75000

-3x^2+150x-10000\\=-3(75000)^2+150(75000)-10000\\=-\$16,863,760,000

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(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

n = 30 -- sample size

\bar x = 50 -- sample mean

\sigma = 10 --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

\sigma_x = \frac{\sigma}{\sqrt n}

\sigma_x = \frac{10}{\sqrt {30}}

\sigma_x = \frac{10}{5.478}

\sigma_x = 1.825

The 95% confidence interval for the z value is:

z = 1.960

Calculate margin of error (E)

E = z * \sigma_x

E = 1.960 * 1.825

E = 3.577

The confidence bound is:

Lower = \bar x - E

Lower = 50 - 3.577

Lower = 46.423

Lower = 46.4 --- approximated

Upper = \bar x + E

Upper = 50 + 3.577

Upper = 53.577

Upper = 53.6 --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

\sigma_x = \frac{\sigma}{\sqrt n}

\sigma_x = \frac{10}{\sqrt {90}}

\sigma_x = \frac{10}{9.49}

\sigma_x = 1.054

The 95% confidence interval for the z value is:

z = 1.960

Calculate margin of error (E)

E = z * \sigma_x

E = 1.960 * 1.054

E = 2.06584

The confidence bound is:

Lower = \bar x - E

Lower = 50 - 2.06584

Lower = 47.93416

Lower = 47.9 --- approximated

Upper = \bar x + E

Upper = 50 + 2.06584

Upper = 52.06584

Upper = 52.1 --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

n = 30     E = 3.577

In (b), we have:

n = 90    E = 2.06584

<em>Notice that the margin of error decreases when the sample size increases.</em>

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