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cluponka [151]
1 year ago
4

Consider following situation, which involves two options. Determine which option is less expensive. Are there unstated factors t

hat might affect your decision?
You currently drive 325 miles per week in a car that gets 14 miles per gallon of gas. You are considering buying a new fuel-efficient car for S16,000 (after trade-in on your current car that gets 47 miles per gallon
premiums for the new and old car are $1000 and $600 per year, respectively. You anticipate spending $1500 per year on repairs for the old car and having no repairs on the new car. Assume gas costs $3.50 pen
ive-year period, is it less expensive to keep your old car or buy the new car?
a five-year period, the cost of the old car iss) and the cost of the new car issThus, over a five-year period, it is less expensive to
and to the nearest dollar as needed
Mathematics
1 answer:
Amiraneli [1.4K]1 year ago
6 0

9514 1404 393

Answer:

  it is less expensive to purchase a new car

Step-by-step explanation:

The expected cost of gas, insurance, and repairs for the old car are ...

  Total miles in 5 years = (325 mi/wk)(52 wk/yr)(5 yr) = 84,500 mi

<u>Old car</u>

  Gas cost in 5 years = (84,500 mi)/(14 mi/gal)($3.50/gal) = $21,125

  Insurance cost in 5 years = (5 yr)($600/yr) = $3,000

  Repair cost in 5 years = ($1500/yr)(5 yr) = $7,500

  Total operating cost in 5 years = $21125 +3000 +7500 = $31,625

__

<u>New Car</u>

  Gas cost in 5 years = (84,500 mi)/(47 mi/gal)($3.50/gal) ≈ $6,293

  Insurance cost in 5 years = (5 yr)($1000/yr) = $5,000

  Repair cost in 5 years = (0)(5 yr) = 0

  Total operating cost in 5 years = $6293 +5000 = 11,293

__

Difference in operating costs is $31,625 -11,293 = $20,332.

It is less expensive to purchase a new car for any price less than $20,300. If the new car costs only $16,000, it is the less-expensive option.

Purchase of a new car is the least-expensive option.

__

<em>Additional comments</em>

The cost of financing, annual license fees, and resale value of the new car may also come into play. In addition, the relative non-economic values of consumption of new materials, adding old materials to the scrap heap, and environmental pollution may figure in the decision. While the car dealer may pay for maintenance costs of a new car, it seems likely that at least one set of tires will be required in 5 years (85000 miles). (Assuming wheel alignment is not an issue for tire wear, that cost may balance out for the two cars.)

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Example 4.5 introduced the concept of time headway in traffic flow and proposed a particular distribution for X 5 the headway be
exis [7]

Answer:

a. k = 3

b. Cumulative distribution function X, F(x)=\left \{ {0} , x\leq 1  \atop {1-x^{-3}, x>1}} \right.

c.  Probability when headway exceeds 2 seconds = 0.125

Probability when headway is between 2 and 3 seconds = 0.088

d. Mean value of headway = 1.5

Standard deviation of headway = 0.866

e.  Probability that headway is within 1 standard deviation of the mean value = 0.9245

Step-by-step explanation:

From the information provided,

Let X be the time headway between two randomly selected consecutive cars (sec).

The known distribution of time headway is,

f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.

a. Value of k.

Since the distribution of X is a valid density function, the total area for density function is unity. That is,

\int\limits^{\infty}_{-\infty} f(x)dx=1

So, the equation becomes,

\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3

b. For this problem, the cumulative distribution function is defined as :

F(x) = \int\limits^1_{\infty} f(x)dx +  \int\limits^x_1 f(x)dx

Now,

F(x) = 0 +  \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 +  \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}

Therefore the cumulative distribution function X is,

F(x)=\left \{ {0} , x\leq 1  \atop {1-x^{-3}, x>1}} \right.

c. Probability when the headway exceeds 2 secs.

Using cdf in part b, the required probability is,

P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125

Probability when headway is between 2 seconds and 3 seconds

Using the cdf in part b, the required probability is,

P(2

≅ 0.088

d. Mean value of headway,

E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5

And,

E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3

The standard deviation of headway is,

= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254

≅ 0.866

e. Probability that headway is within 1 standard deviation of the mean value

P(\alpha - \beta  < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X

From part b, F(x) = 0, if x ≤ 1

=1-(2.366)^{-3}\\=0.9245

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A yard is equal in length to three feet. The function f(x) takes a measurement in yards (as input) and returns a measurement in
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A sea turtle is 3 feet below the surface of the sea. If its position can be recorded as −3 feet, what would the position 0 repre
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Jaleel and Lisa are simplifying the expression 2 (x minus 2) + 2 as shown. Jaleel’s Method 2 (x minus 2) + 2 = 2 x minus 4 + 2 =
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Answer:

(D)Jaleel's method is correct because 2(x-2)=2x-4.

Step-by-step explanation:

Jaleel and Lisa are simplifying the expression 2(x-2)+2 as shown.

J$aleel's Method: \left\{\begin{array}{ccc}2 (x -2) + 2 \\= 2 x - 4 + 2 \\= 2 x - 2\end{array}\right

L$isa's Method: \left\{\begin{array}{ccc}2 (x-2) + 2 \\= 2 x -2 + 2 \\= 2 x\end{array}\right

We can see that Jaleel's method is correct because:

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The correct option is D.

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