Answered

A population of bacteria is 8500 on Day 1, 9350 on Day 2, and 10285 on Day 3.


What will the population be on Day 30?

A11314
B33150
C134836
D148320

Verified Answer :

InesWalston

Answer:

The population on 30th day will be 134836.

Step-by-step explanation:

This is a case of exponential growth. The general form of exponential function is,

\(y=ab^x\)

where, a and b are constants.

The data points from the question are \((1,8500),(2,9350),(3,10285)\)

Putting the values in the function,

for \(x=1, y=8500\)

\(8500=ab\)  ----------1

for \(x=2, y=9350\)

\(9350=ab^2\)  ------2

Dividing equation 2 by 1,

\(\Rightarrow \dfrac{ab^2}{ab}=\dfrac{9350}{8500}=1.1\)

\(\Rightarrow b=1.1\)

Putting the value of b in equation 1,

\(\Rightarrow a=\dfrac{8500}{1.1}=7727.27\)

Now the function becomes,

\(y=7727.27(1.1)^x\)

Putting x=30, we get

\(y=7727.27(1.1)^{30}=134836.2\approx 134836\)

JeanaShupp

Answer: 134836

Step-by-step explanation:

The exponential growth equation is given by :-

\(y=Ab^x\)                              (1)

, where A= initial population

b= multiplicative growth arte.

x= Time period.

Given : A population of bacteria is 8500 on Day 1, 9350 on Day 2, and 10285 on Day 3.

i.e. A= 8500

\(b=\dfrac{9350}{8500}=1.1\)  [Multiplicative growth rate is the common ration between two terms.]

Put value of A and b in (1) , we get

\(y=8500(1.1)^x\)                     (2)

For Day 30

When x= 30-1=29            [∵Day 1 is the initial day.]

Put x= 29 in (2), we get

\(y=8500(1.1)^{29}=8500(15.8630929717)\\\\=134836.290259approx134836\)        [Round to the nearest whole number.]

Hence, the population be on Day 30 will be 134836 .

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