Question

SAT Math Help (MANY THANKS!)

Answer 1

Answer:

48.8%

Step-by-step explanation:

So let's start at the temperature $59^\circ C$ where maximum reproduction is achieved.  Let's say the maximum reproduction rate number is $x$.

The problem says the reproduction rate is reduced by 20 percent per decrease in degree of temperature in Celsius.

So a $58^\circ C$, the reproduction number is $x-.2(x)$ since we are taking 20% of the number we began with.

Let's simplify $x-.2(x)$.

These are like terms; 1-.2 is .8 so $x-.2x=.8x$

Now let's go to $57^\circ C$, the reproduction rate is decreased by 20% from the .8x. So we are taking 20% off of .8x:

$.8x-.2(.8x)$

$.8x-.16x$

$.64x$

Going to finally $56^\circ C$, the reproduction rate is decreased by 20% from .64x. So we are taking 20% off of .64x:

$.64x-.2(.64x)$

$.64x-.128x$

$.512x$

So that is 51.2% of x.

So it was reduce by 100%-51.2% =48.8% .

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Another way if you are good with exponential functions:

$A=P_0(1-.2)^t$

The initial reproduction rate is $P_0$.

Let 59 degree C represent what happens at time t=0.

So 58 degree C represents what happens at time t=1.

...

56 degree C represents what happens at time t=3.

Plug in t=3:

$A=P_0(1-.2)^3$

$A=P_0(.8)^3$

$A=P_0(.512)$

This says at time t=3 that you have 51.2% of your reproduction rate which means it was decreased by 100%-51.2%=48.8% from it's original reproduction rate at t=0 which was $59^\circ C$.