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avanturin [10]
3 years ago
9

A.Suntrust Bank approved Bob for a mortgage loan of $135,000. In order for him to receive the funds, he must make a down payment

of 18%. How much money must he have for the down payment? Be sure to include the dollar sign in your answer
B. One person takes up about 3 square feet of space. Use this value to estimate how many people can fit in an area that is 290 feet wide and 230 feet long.


C. A storage closet is a rectangle, and has a length of 15 feet, width of 10 feet, and a height of 9 feet. Joe wants to store boxes in the closet. Each box has a length of 2 feet, a width of 1.5 feet, and a height of 1 foot. How many boxes can he ideally fit into the closet?




it's a 3 Part Question
Mathematics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

\boxed{\text{A. \$24 300; B. 22 333 people; C. 450 boxes}}

Step-by-step explanation:

A. Sunkist Bank

To get 18 % of a number, multiply by ¹⁸/₁₀₀ or 0.18.

\text{Down payment} = \$135 000 \times 0.18 = \mathbf{\$24 300}\\\text{Bob's down payment is } \boxed{\textbf{\$24 300}}

B. Number of people

(1) Calculate the area

A = lw = 290 ft × 230 ft = 66 700 ft²

(2) Calculate the number of people

You can fit 1 person into 3 ft² of space.

\text{Number of people} = \text{66 700 ft}^{2} \times \dfrac{\text{1 person}}{\text{300 ft}^{2}} = \textbf{22 333 people}\\\\\text{The space will hold $\boxed{\textbf{22 333 people}}$}

C.  Number of boxes

(1) Volume of closet

V = lwh = 15 ft × 10 ft × 9 ft = 1350 ft³

(2) Volume of a box

V = lwh = 2 ft × 1.5 ft × 1 ft = 3 ft³

(3) Number of boxes

\text{Number of boxes} = \text{1350 ft}^{3} \times \dfrac{\text{1 box }}{\text{3 ft}^{3}} = \textbf{450 boxes}\\\\\text{Ideally, Bob can fit $\boxed{\textbf{450 boxes}}$ into the closet}

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(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

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\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

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0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

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W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

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-0.02 + 0.00004R = 0

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Make R the subject

R = \frac{0.02}{0.00004}

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When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

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Collect like terms

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Solve for W

W = \frac{-0.07875}{ - 0.001}

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When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

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Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

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-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

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So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

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