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
Function: $f(t) = 1100(1.0905)^t$ Growth: $9.05$ % increase per hour
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
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
The initial number of bacteria $P_0 = 1100$.
Since bacteria double every 8 hours, solve for growth factor $b$:
$$2 = b^8 \implies b = 2^{\frac{1}{8}}$$
Calculate $2^{\frac{1}{8}} \approx 1.0905$.
Use the form $f(t) = P_0 b^t$:
$$f(t) = 1100(1.0905)^t$$
Subtract 1 from the growth factor and convert to percentage:
$$(1.0905 - 1) \times 100 = 9.05\%$$
Function: $f(t) = 1100(1.0905)^t$
Growth: $9.05$ % increase per hour
The initial number of bacteria $P_0 = 1100$.
Since bacteria double every 8 hours, solve for growth factor $b$:
$$2 = b^8 \implies b = 2^{\frac{1}{8}}$$
Calculate $2^{\frac{1}{8}} \approx 1.0905$.
Use the form $f(t) = P_0 b^t$:
$$f(t) = 1100(1.0905)^t$$
Subtract 1 from the growth factor and convert to percentage:
$$(1.0905 - 1) \times 100 = 9.05\%$$
Function: $f(t) = 1100(1.0905)^t$
Growth: $9.05$ % increase per hour
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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