Question 3 If we have 1 _E. COLL bacteria , at the beginning I, in . 5 HOUR S time how many E .coli we will have? a 1.6times 104(1.6times 10 to the power 4) b 3.2times 104(3.2times 10 to the power 4) C 6.4times 104(6.4times 10 to the power 4) d 8.1times 103(8.1times 10 to the power 3) Previous page

To solve this problem, we need to understand the growth pattern of E. coli bacteria. E. coli typically doubles every 20 minutes under optimal conditions. Given that we 1 bacterium, we can calculate the number of bacteria after 5 hours by determining how many 20-minute intervals fit into 5 hours and then doubling the initial amount that many times.<br /><br />First, let's convert 5 hours into minutes:<br />\[ 5 \text{ hours} \times 60 \text{ minutes/hour} = 300 \text{ minutes} \]<br /><br />Next, we determine how many 20-minute intervals are in 300 minutes:<br />\[ \frac{300 \text{ minutes}}{20 \text{ minutes/interval}} = 15 \text{ intervals} \]<br /><br />Since the bacteria double every interval, we can use the formula for exponential growth:<br />\[ \text{Final amount} = \text{Initial amount} \times 2^{\text{number of intervals}} \]<br />\[ \text{Final amount} = 1 \times 2^{15} \]<br /><br />Calculating \(2^{15}\):<br />\[ 2^{15} = 32768 \]<br /><br />So, the final amount of E. coli bacteria after 5 hours is:<br />\[ 32768 \]<br /><br />Now, let's express this in scientific notation:<br />\[ 32768 = 3.26768 \times 10^4 \]<br /><br />Rounding to the nearest option given:<br />\[ 3.2 \times 10^4 \]<br /><br />Therefore, the correct answer is:<br />b) \(3.2 \times 10^4\)