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3 years ago

Algebraically find three consecutive positive integers such that the square of the first is equal to the third.

1 answer:

7 0

Your numbers are 2, 3, 4.

The three consecutive numbers are x, x+1, x+2
The square of the first equals the third so x^2=x+2
Solve this quadratic as x^2-x-2=0
Factor into (x+1)(x-2)=0
So your answers are x=-1 and x=2
Since the question says they’re positive integers, use x=2
So the numbers are 2, 3, 4

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3 years ago

Anna's math test grade was 7 points less than her science test grade. The average grade was 91.5%. What did she get on each test

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84.5%

Step-by-step explanation:

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2 years ago

The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she d

aksik [14]

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)

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

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

The manager wants to give free hamburgers to only 1% of her costumers, which means that probability of a costumer getting a free hamburger is 0.01

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

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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2 years ago