Answer:
![y=0.1(2)^x](https://img.qammunity.org/2023/formulas/mathematics/high-school/r74bjebwnc46d0i8b0o6gqtn358xl5iu9t.png)
where x is time in hours, and y is the number of bacteria in percent in decimal form.
Explanation:
General form of an exponential function:
![y=ab^x](https://img.qammunity.org/2023/formulas/mathematics/high-school/hye5rg1h8wj3ohgdt4j1vpepdhoym0w9ex.png)
where:
- a is the initial value (y-intercept).
- b is the base (growth/decay factor) in decimal form.
If b > 1 then it is an increasing function.
If 0 < b < 1 then it is a decreasing function. - x is the independent variable.
- y is the dependent variable.
Define the variables:
- Let x = time (in hours)
- Let y = number of bacteria (in percent in decimal form)
If the bacteria grows on food by doubling every hour then the growth rate b = 2.
If the bacteria initially covers 10% of the food then y = 10% when x = 0.
Therefore, a = 10% = 0.1.
Therefore, the equation is:
![y=0.1(2)^x](https://img.qammunity.org/2023/formulas/mathematics/high-school/r74bjebwnc46d0i8b0o6gqtn358xl5iu9t.png)
Substitute x = 3.3 into the found equation:
![\begin{aligned}x=3.3 \implies y & =0.1(2)^(3.3)\\& =0.9849155307...\\& =1.0\:\sf (nearest\:tenth)\end{aligned}](https://img.qammunity.org/2023/formulas/mathematics/high-school/pobnww4hh301cecpv8z6lfq8w022yb20nx.png)
As 1.0 = 100%, this justifies the claim.
![\begin{array}c\cline{1-3} x & y & y \\\cline{1-3} 0 & 10\% & 0.1\\\cline{1-3} 1 & 20\% & 0.2\\\cline{1-3} 2 & 40\% & 0.4\\\cline{1-3} 3 & 80\%\ & 0.8\\\cline{1-3} 4 & 160\% & 1.6\\\cline{1-3}\end{array}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ga0wlsotwaqm5r5zgzdyvw8qwis0osmtmm.png)