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lianna [129]
3 years ago
10

Find a linear equation to model this real-world application:

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
3 0

The linear equation to model the company's monthly expenses is y = 2.5x + 3650

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

Let "x" be the units produced in a month

It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers.

Cost per unit = $ 2.50

The company has monthly operating expenses of $350 for utilities and $3300 for salaries

We have to write the linear equation

The linear equation to model the company's monthly expenses in the form of:

y = mx + b

Cost per unit = $ 2.50

Monthly Expenses = $ 350 for utilities and $ 3300 for salaries

Let "y" be the total monthly expenses per month

Then,

Total expenses = Cost per unit(number of units) + Monthly Expenses

y = 2.50(x) + 350 + 3300\\\\y = 2.5x + 3650

Thus the linear equation to model the company's monthly expenses is y = 2.5x + 3650

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<span><u>Answer </u>
31.12 in

<u>Explanation </u>
To get the length required, lets first find angle mF,
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DE/sin75=22/sin⁡43
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Answer:

Approximately 0.15 (360 / 2401.) (Assume that the choices of the 5 passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all 7 floors.)

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If there is no requirement that no two passengers exit at the same floor, each of these 5 passenger could choose from any one of the 7 floors. There would be a total of 7 \times 7 \times 7 \times 7 \times 7 = 7^{5} unique ways for these 5\! passengers to exit the elevator.

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The first passenger could choose from any of the 7 floors.

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Likewise, the third passenger would have to choose from only (7 - 2) = 5 floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only (7 \times 6 \times 5 \times 4 \times 3) unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the 7 floors. Each of these 7^{5} combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}.

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Step-by-step explanation:

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