sarahkiser16
contestada

Jane is growing two kinds of bacteria in her lab. She starts with 300 bacteria of sample 1 in a petri dish and they multiply at a rate of 5 percent per minute. Five minutes later, she starts growing sample 2 with the same starting amount of bacteria. These bacteria multiply at a rate of 8 percent per minute. Which inequality can be used to find how many minutes it will take until the number of bacteria in sample 2 exceeds the number of bacteria in sample 1?

A300(1.08)^t>300(1.05)^t , B 300(1.08)^t-5>300(1.05)^t, C300(1.05)^t>300(1.08)^t-5, D300(0.92)^t-5>300(0.92)^t, E300(0.92)^t-5>300(0.92)^t-5

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Respuesta :

nullgaming650
I Believe The Answer Is:

 B 300(1.08)^t-5>300(1.05)^t

Hope I Helped :D

-Nullgaming650
JeanaShupp

Answer: [tex]\text{B. } 300(1.08)^{t-5}>300(1.05)^t[/tex]

Step-by-step explanation:

The exponential growth function with rate of growth 'r' in time 't' years is given by :-

[tex]f(t)=A(1+r)^t[/tex], where A is the initial amount.

Given: For Sample 1 :

Initial amount of bacteria = 300

Rate of growth = 5%=0.05

Let t be the time period.

The exponential function that models the growth of bacteria in sample 1 :-

[tex]f(t)=300(1+0.05)^t=300(1.05)^t[/tex]

For Sample 2 :

Initial amount of bacteria = 300

Rate of growth = 8%=0.08

Time period = t-5

The exponential function that models the growth of bacteria in sample 2 :-

[tex]g(t)=300(1+0.08)^{t-5}=300(1.05)^{t-5}[/tex]

The inequality which can be used to find how many minutes it will take until the number of bacteria in sample 2 exceeds the number of bacteria in sample 1 is given by :-

[tex]300(1.08)^{t-5}>300(1.08)^t[/tex]

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