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in a lab experiment, 1,100 bacteria are placed in a petri dish. the con…

Function: $f(t) = 1100(1.0905)^t$ Growth: $9.05$ % increase per hour

Categoría: physics Actualizado: 2026-02-09

Pregunta

in a lab experiment, 1,100 bacteria are placed in a petri dish. the conditions are such that the number of bacteria is able to double every 8 hours. write a function showing the number of bacteria after $t$ hours, where the hourly growth rate can be found from a constant in the function. round all coefficients in the function to four decimal places. also, determine the percentage of growth per hour, to the nearest hundredth of a percent.
answer
function: $f(t) = \\square(\\square)^\\square$
growth $\\square$ % increase per hour

Pasos de solución

  1. Understand the question

    in a lab experiment, 1,100 bacteria are placed in a petri dish. the conditions are such that the number of bacteria is able to double every 8 hours. write a function showing the number of bacteria after $t$ hours, where the hourly growth rate can be found from a constant in the function. round all coefficients in the function to four decimal places. also, determine the percentage of growth per hour, to the nearest hundredth of a percent.
    answer
    function: $f(t) = \\square(\\square)^\\square$
    growth $\\square$ % increase per hour

  2. Explanation

    Step1: Identify initial bacteria count

    The initial number of bacteria $P_0 = 1100$.

    Step2: Find hourly growth factor

    Since bacteria double every 8 hours, solve for growth factor $b$:
    $$2 = b^8 \implies b = 2^{\frac{1}{8}}$$
    Calculate $2^{\frac{1}{8}} \approx 1.0905$.

    Step3: Write exponential growth function

    Use the form $f(t) = P_0 b^t$:
    $$f(t) = 1100(1.0905)^t$$

    Step4: Calculate hourly growth percentage

    Subtract 1 from the growth factor and convert to percentage:
    $$(1.0905 - 1) \times 100 = 9.05\%$$

  3. Final answer

    Function: $f(t) = 1100(1.0905)^t$
    Growth: $9.05$ % increase per hour

Respuesta

Explanation

Step1: Identify initial bacteria count

The initial number of bacteria $P_0 = 1100$.

Step2: Find hourly growth factor

Since bacteria double every 8 hours, solve for growth factor $b$:
$$2 = b^8 \implies b = 2^{\frac{1}{8}}$$
Calculate $2^{\frac{1}{8}} \approx 1.0905$.

Step3: Write exponential growth function

Use the form $f(t) = P_0 b^t$:
$$f(t) = 1100(1.0905)^t$$

Step4: Calculate hourly growth percentage

Subtract 1 from the growth factor and convert to percentage:
$$(1.0905 - 1) \times 100 = 9.05\%$$

Answer

Function: $f(t) = 1100(1.0905)^t$
Growth: $9.05$ % increase per hour

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Question Analysis

Subject natural science
Sub Subject biology
Education Level high school
Difficulty unspecified
Question Type calculation
Multi Question No
Question Count 1
Analysis Status completed
Analyzed At 2026-02-09T20:28:03

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in a lab experiment, 1,100 bacteria are placed in a petri dish. the conditions are such that the number of bacteria is able to double every 8 hours. write a function showing the number of bacteria after $t$ hours, where the hourly growth rate can be found from a constant in the function. round all coefficients in the function to four decimal places. also, determine the percentage of growth per hour, to the nearest hundredth of a percent.
answer
function: $f(t) = \\square(\\square)^\\square$
growth  $\\square$ % increase per hour

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natural sciencebiologycalculationhigh schoolturns-1

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