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Vikki [24]
3 years ago
9

It takes 60 minutes to make 2000 copies on an old copy machine. When a newer copier works alongside the old copier, 2000 copies

can be made in 15 minutes. How many minutes are required for the newer copier working alone to make 2000 copies?
Enter your answer in the box

_____ min

CAN ANYONE HELP ME PLEASE ASAP. NEED HELP REALLY BADLY!!!!
Mathematics
2 answers:
kramer3 years ago
7 0
20 minutes is your answer i believe
zheka24 [161]3 years ago
3 0

Answer:

20\ minutes

Step-by-step explanation:

we know that

An old copy machine takes 60 minutes to make 2,000  copies

so

by proportion

Find how many copies makes in 15 minutes

\frac{2,000}{60}=\frac{x}{15}

x=2,000*15/60=500\ copies

Remember that

A newer copier works alongside the old copier takes 15 minutes to make 2,000  copies

that means that the newer copier in 15 minutes makes 2,000-500=1,500  copies

so

The newer copier takes 15 minutes to make 1,500  copies

by proportion

Find how many minutes are required for the newer copier working alone to make 2,000 copies

\frac{15}{1,500}=\frac{x}{2,000} \\ x=2,000*15/1,500\\x=20\ minutes

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B.

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Step-by-step explanation:

The bacteria culture starts with 500 bacteria and doubles size every half hour.= 2 hours

32,000

Thus, after three hours, the population of bacteria is 32,000.= 3 hours

A bacteria culture starts with 500 bacteria and doubles in size every half hour.1

(a) How many bacteria are there after 3 hours?

We are told “. . . doubles in size every half hour.” Let’s make a table of the time and population:

t 0 0.5 1.0 1.5 2.0 2.5 3.0

bacteria 500 1000 2000 4000 8000 16000 32000

Thus, after three hours, the population of bacteria is 32,000.

(b) How many bacteria are there after t hours?

In t hours, there are 2t doubling periods. (For example, after 4 hours, the population has doubled 8

times.) The initial value is 500, so the population P at time t is given by

P(t) = 500 · 2

2t

This is an acceptable response, but in calculus and all advanced mathematics and science, we will

almost always want to use the natural exponential base, e. Let’s redo the problem using the natural

exponential growth function

P(t) = P0e

rt

We are given that the initial population is 500 bacteria. So P0 = 500 and we have

P(t) = 500ert

We know that after 1 hour, there are 2000 bacteria. (We could’ve used any other pair from our table

that we wish.) Substituting into our function, we get

P(t) = 500ert

2000 = 500er·1

2000 = 500er

1

500

· 2000 =

1

500

· 500er

4 = er

and now we use the natural logarithm to solve for r

ln (4) = ln (er

)

ln (4) = r · ln (e)

ln (4) = r

1.3863 ≈ r

Thus the function P(t) = 500e1.3863t gives the number of bacteria after t hours.

(c) How many bacteria are there after 40 minutes?

The input t to our function is given in hours, so we must convert 40 minutes into hours. So (40 min)

1 hr

60 min

=

2

3

hr.

P(t) = 500e1.3863t

P

2

3

= 500e1.3863·(2/3)

P

2

3

≈ 1259.9258

Thus, after 40 minutes (2/3 of an hour), there are about 1260 bacteria.

(d) Graph the population function and estimate the time for the population to reach 100,000.

50000

100000

1 2 3 4 5

b

(3.8219, 100,000)

t

P

Using the calc:intersect on the TI-84, we get the point (3.8219, 100000), thus the population will

reach 100,000 in about 3.82 yrs

6 0
3 years ago
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