What is the simplified version of this expression 18/4(8/9b+8/6)?

Mathematics

1 answer:

Vadim26 [7]3 years ago

6 0

18/4(8/9b+8/6)
=4b+6
hope this helps :)

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Suppose you want to buy a new car and are trying to choose between two models: Model A: costs $16,500 and its gas mileage is 25

laila [671]

Answer:

1. CA=16,500+5,050x

2. CB=24,500+3,450x

3. CA(x=5)=CB(x=5)=41,750

If keeped 4 years, Model A is more economical.

4. 5 years

5. From month 49 to 61.

6. The cost of ownership of Model A increases more than Model B, as it is less gas efficient. The break-even point for x is reduced from x=5 to x=2.35.

7. The fixed cost are reduced by a 40%, so the variable cost, the ones that depend on time of ownership, are increased in importance.

Step-by-step explanation:

We can express the cost of ownership as the sum of the purchase cost, gas cost and insurance cost.

1. For model A we have:

$\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$16,500+3 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{25\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$250\cdot x\\\\\\\text{Cost of ownership}=\$16,500+\$4,800x+\$250x\\\\\\\text{Cost of ownership}=$16,500+\$5,050x$

2. For model B we have:

$\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$24,500+3 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{40\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$450\cdot x\\\\\\\text{Cost of ownership}=\$24,500+\$3,000x+\$450x\\\\\\\text{Cost of ownership}=$24,500+\$3,450x$

3. If x=5, the costs for each car are:

$\text{CoOwn A}=16,500+5,050\cdot(5)=16,500+25,250=41,750\\\\\\\text{CoOwn B}=24,500+3,450\cdot(5)=24,500+17,250=41,750$

5 years is the break-even point for the cost of ownership between these two cars.

If you plan to keep the car for 4 years, the costs are:

$\text{CoOwn A}=16,500+5,050\cdot(4)=16,500+20,200=36,700\\\\\\\text{CoOwn B}=24,500+3,450\cdot(4)=24,500+13,800=38,300$

For a 4 year period ownership, the model A is more economical ($36,700).

4. This happens for 1 year, the fifth year, in which the two models have the same cost of ownership.

5. At the 5th year, the cost for both models are the same.

Then, this corresponds to the months 4*12+1=48+1=49 and 5*12+1=61.

6. If the cost of gas doubles, the cost of ownership would rise for both model, but more for the Model A, which is less gas efficient and hence has a higher gas cost.

Model A

$\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$16,500+6 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{25\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$250\cdot x\\\\\\\text{Cost of ownership}=\$16,500+\$9,600x+\$250x\\\\\\\text{Cost of ownership}=$16,500+\$9,850x$

Model B

$\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$24,500+6 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{40\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$450\cdot x\\\\\\\text{Cost of ownership}=\$24,500+\$6,000x+\$450x\\\\\\\text{Cost of ownership}=$24,500+\$6,450x$