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BigorU [14]
3 years ago
15

Monica reads 7 1/2 pages of a book in 9 minutes. What is her average reading rate in pages per minute?

Mathematics
2 answers:
Schach [20]3 years ago
5 0

Answer:

Monica can read \frac{5}{6} pages per minute.

Step-by-step explanation:

Monica reads 7 1/2 pages of a book in 9 minutes.

Or Monica reads \frac{15}{2} pages of a book in 9 minutes.

So, per minute she can read \frac{\frac{15}{2}}{9} pages

= \frac{15}{2}\times\frac{1}{9} =\frac{5}{6} pages

Hence, Monica can read \frac{5}{6} pages per minute.

Assoli18 [71]3 years ago
3 0
\frac{7.5}{9} =  \frac{n}{1}

or in other words divide 7.5 by 9 to get your answer which is .83333.. which is the same as 5/6 of a page per minute.

Check
5/6 + 5/6 + 5/6 + 5/6 + 5/6 +5/6 +5/6 +5/6 +5/6 + 45/6 or 7 3/6 = 7 1/2
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Answer:

The number of minutes advertisement should use is found.

x ≅ 12 mins

Step-by-step explanation:

(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

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Where Average waiting time = μ = 2.5

The function f(t) becomes

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<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

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<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>

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