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3 years ago

You take your favorite teacher out to lunch. The total

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

7 0

Answer:

it would be 45 total with the 20%

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A culture of bacteria has an initial population of 9300 bacteria and doubles every 3

sp2606 [1]

Answer:

The approximate population of bacteria in the culture after 10 hours is 93,738.

Step-by-step explanation:

<h3>General Concepts:</h3>

Exponential Functions.
Exponential Growth.
Doubling Time Model.
Logarithmic Form.

BPEMDAS Order of Operations:

Brackets.
Parenthesis.
Exponents.
Multiplication.
Division.
Addition.
Subtraction.

<h2>Definitions:</h2>

We are given the following Exponential Growth Function (Doubling Time Model), $\displaystyle\mathsf{P_{(t)}\:=\:P_0\cdot2^{(t/d)}}$ where:

$\displaystyle\sf{P_t\:\:\rightarrow}$ The population of bacteria after “<em>t </em>” number of hours.
$\displaystyle\sf{P_0 \:\:\rightarrow}$ The initial population of bacteria.
$\displaystyle{t \:\:\rightarrow}$ Time unit (in hours).
$\displaystyle{\textit d \:\:\rightarrow}$ Doubling time, which represents the amount of time it takes for the population of bacteria to grow exponentially to become twice its initial quantity.

<h2>Solution:</h2>

<u>Step 1: Identify the given values.</u>

$\displaystyle\sf{P_0\:=}$ 9,300.
<em>t</em> = 10 hours.
<em>d</em> = 3.

<u>Step 2: Find value.</u>

1. Substitute the values into the given exponential function.

$\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}$

$\displaystyle\mathsf{\longrightarrow P_{(10)} = 9300\cdot2^{(10/3)}}$

2. Evaluate using the BPEMDAS order of operations.

$\displaystyle\mathsf{P_{(10)} = 9300\cdot2^{(10/3)}\quad \Longrightarrow BPEMDAS:\:(Parenthesis\:\:and\:\:Division).}$

$\displaystyle\sf P_{(10)} = 9300\cdot2^{(3.333333)}\quad\Longrightarrow BPEMDAS:\:(Exponent).}$

$\displaystyle\sf P_{(10)} = 9300\cdot(10.079368399)\quad \Longrightarrow BPEMDAS:(Multiplication).}$

$\boxed{\displaystyle\mathsf{P_{(10)} \approx 93,738.13\:\:\:or\:\:93,738}}$

Hence, the population of bacteria in the culture after 10 hours is approximately 93,738.

<h2>Double-check:</h2>

We can solve for the amount of <u>time</u> <u>(</u><em>t</em> ) it takes for the population of bacteria to increase to 93,738.

1. Identify given:

$\displaystyle\mathsf{P_{(t)} = 93,738 }$ .
$\displaystyle\mathsf{P_0 = 9,300}$ .
<em>d </em>= 3.

2. Substitute the values into the given exponential function.

$\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}$

$\displaystyle\mathsf{\longrightarrow 93,378 = 9,300\cdot2^{(t/3)}}$

3. Divide both sides by 9,300:

$\displaystyle\mathsf{\longrightarrow \frac{93,378}{9,300} = \frac{9,300\cdot2^{(t/3)}}{9,300}}$

$\displaystyle\mathsf{\longrightarrow 10.07936840 = 2^{(t/3)}}$

4. Transform the right-hand side of the equation into logarithmic form.

$\boxed{\displaystyle\mathsf{\underbrace{ x = a^y}_{Exponential\:Form} \longrightarrow \underbrace{y = log_a x}_{Logarithmic\:Form}}}$

$\displaystyle\mathsf{\longrightarrow 10.07936840 = \bigg[\:\frac{t}{3}\:\bigg]log(2)}$

5. Take the <em>log</em> of both sides of the equation (without rounding off any digits).

$\displaystyle\mathsf{log(10.07936840) = \bigg[\:\frac{t}{3}\:\bigg]log(2)}$

$\displaystyle\mathsf{\longrightarrow 1.003433319 = \bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996)}$

6. Divide both sides by (0.301029996).

$\displaystyle\mathsf{\frac{1.003433319}{0.301029996} = \frac{\bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996) }{0.301029996}}$

$\displaystyle\mathsf{\longrightarrow 3.3333333 = \frac{t}{3}}$

7. Multiply both sides of the equation by 3 to isolate "<em>t</em>."

$\displaystyle\mathsf{(3)\cdot(3.3333333) = \bigg[\:\frac{t}{3}\:\bigg]\cdot(3)}$

$\boxed{\displaystyle\mathsf{t\approx10}}$

Hence, it will take about 10 hours for the population of bacteria to increase to 93,378.

__________________________________

Learn more about Exponential Functions on:

brainly.com/question/18522519

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