I11)
The growth of bacteria makes it necessary to time-date some food products so that they will be sold and consum
present is given by , where is time in days and the value of is in millions. Find the number of ba
(a) 3 days (b) 4 days © 2 weeks
(a) Approximately million bacteria are present in 3 days.
(Round to the nearest integer as needed.)
(b) Approximately million bacteria are present in 4 days.
(Round to the nearest integer as needed.)
© Approximately million bacteria are present in 2 weeks.
(Round to the nearest integer as needed.)

Ask by Washington Bird. in the United States

Nov 18,2024

Question

I11)
The growth of bacteria makes it necessary to time-date some food products so that they will be sold and consum
present is given by , where is time in days and the value of is in millions. Find the number of ba
(a) 3 days (b) 4 days © 2 weeks
(a) Approximately million bacteria are present in 3 days.
(Round to the nearest integer as needed.)
(b) Approximately million bacteria are present in 4 days.
(Round to the nearest integer as needed.)
© Approximately million bacteria are present in 2 weeks.
(Round to the nearest integer as needed.)

Ask by Washington Bird. in the United States

Nov 18,2024

UpStudy AI · Accepted Answer

Given the growth function of bacteria as $$ f(t) = 600 e^{0.1t} $$, where $$ t $$ is time in days and the value of $$ f(t) $$ is in millions.

To find the number of bacteria present at different time intervals, we need to substitute the given values of time into the function and calculate the result.

(a) For 3 days:
Substitute $$ t = 3 $$ into the function:
$$ f(3) = 600 e^{0.1 	imes 3} $$

(b) For 4 days:
Substitute $$ t = 4 $$ into the function:
$$ f(4) = 600 e^{0.1 	imes 4} $$

(c) For 2 weeks (14 days):
Substitute $$ t = 14 $$ into the function:
$$ f(14) = 600 e^{0.1 	imes 14} $$

Let's calculate the values for each case.
Calculate the value by following steps:
- step0: Calculate:
  $$600e^{0.1	imes 3}$$
- step1: Multiply the numbers:
  $$600e^{0.3}$$
- step2: Convert the expressions:
  $$600e^{\frac{3}{10}}$$
- step3: Rewrite the expression:
  $$600\sqrt[10]{e^{3}}$$
For 3 days, approximately 809.9 million bacteria are present.

Now, let's calculate the number of bacteria present in 4 days.
Calculate the value by following steps:
- step0: Calculate:
  $$600e^{0.1	imes 4}$$
- step1: Multiply the numbers:
  $$600e^{0.4}$$
- step2: Convert the expressions:
  $$600e^{\frac{2}{5}}$$
- step3: Rewrite the expression:
  $$600\sqrt[5]{e^{2}}$$
For 4 days, approximately 895.1 million bacteria are present.

Next, let's calculate the number of bacteria present in 2 weeks (14 days).
Calculate the value by following steps:
- step0: Calculate:
  $$600e^{0.1	imes 14}$$
- step1: Multiply the numbers:
  $$600e^{1.4}$$
- step2: Convert the expressions:
  $$600e^{\frac{7}{5}}$$
For 2 weeks (14 days), approximately 2433.1 million bacteria are present.

Therefore, the number of bacteria present at different time intervals are:
(a) Approximately 809.9 million bacteria are present in 3 days.
(b) Approximately 895.1 million bacteria are present in 4 days.
(c) Approximately 2433.1 million bacteria are present in 2 weeks.
(a) Approximately 810 million bacteria are present in 3 days. (b) Approximately 895 million bacteria are present in 4 days. (c) Approximately 2433 million bacteria are present in 2 weeks.

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