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saw5 [17]
3 years ago
14

a water turck is filling a swiming pool. the euqation that represents this relationship is y = 19.75x where y is the number of g

allons of water in the pool and x is the number of minutes the truck has been filling the pool choose true or false for each statement
Mathematics
1 answer:
Levart [38]3 years ago
8 0

Answer:

<h2>True </h2>

Step-by-step explanation:

Given the expression

y = 19.75x

let us see the solution from this angle

we know that

speed=distance/time

distance=speed*time

i.e quantity=rate*time

y = 19.75x

y=the quantity =number of gallons

19.5= the rate of flow

therefore

x= the time in  minutes

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The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti
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Answer:

1) 20%

2) Choice a.

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P(t)=10000(0.2)^t

1) P(0) is the population initially.

P(1) is the population after a year.

\frac{P(1)}{P(0)} represents the population increase factor.

So let's evaluate that fraction:

\frac{P(1)}{P(0)}

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\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2

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2) Let's figure out the population growth in terms of months instead of years.

P(t)=10000(0.2)^{t}

We want t to represent months.

A full year is 12 months, in a full year we have that P(1)=10000(0.2)^1=10000(0.2)=2000

So we want a new P such that P(12)=2000 since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) P(12)=10000(0.87449)^{12}=2000 \text{approximately}

b) P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}

c) P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}

d) P(12)=10000(0.87449)^{12+12}=400 \text{approximately}

So a is the function we want.

Also another way to look at this:

P(t)=10000(.2)^t where t is in years.

P(t)=10000(.2^\frac{1}{12})^t where t is in months.

And .2^\frac{1}{12}=0.874485 \text{approximately}

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