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marishachu [46]

3 years ago

9

A car recently sold for $32,345. if there's a 6% sales tax on automobile sales, how much tax will be added to the price of the c

ar?
a. $941.07
b. $194.70
c. $1,940.70
d. $34,285.70

2 answers:

grandymaker [24]3 years ago

8 0

The correct answer to this question is C

EastWind [94]3 years ago

4 0

<span>$1,940.70 will be added to the price of the car. </span>
<span>$32,345 * 6/100 = $1,940.70</span>

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x ≅ 12 mins

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(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

<h3 /><h3>Step 1</h3>

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

$f(t)=\left \{ {{0 ,\-t$

Consider the second function:

$f(t)=\frac{e^{-t/\mu}}{\mu}\\$

Where Average waiting time = μ = 2.5

The function f(t) becomes

$f(t)=0.4e^{-0.4t}$

<h3>Step 2</h3>

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The probability that a costumer has to wait for more than x minutes is:

$\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt$

which is equal to 0.01

<h3>Step 3</h3>

Solve the equation for x

$\int\limits^{\infty}_x {0.4e^{-0.4t}} \, dt =0.01\\\\\frac{0.4e^{-0.4t}}{-0.4}=0.01\\\\-e^{-0.4t} |^\infty_x =0.01\\\\e^{-0.4x}=0.01$

Take natural log on both sides

$ln (e^{-0.4x})=ln(0.01)\\-0.4x=ln(0.01)\\-0.4x=-4.61\\x= 11.53$

<h3>Results</h3>

The costumer has to wait x = 11.53 mins ≅ 12 mins to get a free hamburger

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