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Sunny_sXe [5.5K]
2 years ago
5

For spring break you and some friends plan a road trip to a sunny destination that is 1705 miles away. if you drive a car that g

ets 39 miles per gallon and gas costs $3.339/gal, about how much will it cost to get to your destination?
Mathematics
2 answers:
goldfiish [28.3K]2 years ago
6 0

Answer:

$145.98

Step-by-step explanation:

You and some friends plan a road trip to a sunny destination that is 1705 miles away.

You drive a car that gets 39 miles per gallon.

Total gas used in the trip = \frac{1705}{39}

                                          = 43.7179487 ≈ 43.72 gallons

Gas costs per gallon = $3.339

Total cost to get to your destination = 43.72 × 3.339

                                                            = $145.98108 ≈ $145.98

The cost would be $145.98 to get to your destination.

STatiana [176]2 years ago
5 0
1705/39 = 43.718 gallons
43.718*3.339 = $145.97 
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Answer:

a. k = 3

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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.

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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

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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}

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

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Probability when headway is between 2 seconds and 3 seconds

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

P(2

≅ 0.088

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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

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

≅ 0.866

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

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