A culture of bacteria obeys the law of uninhibited growth. if 500 bacteria are present initially, and there are 800 after 1 hour , how many will be present after 5 hours? round to the nearest whole number.
2 answers:
Since this culture of bacteria obeys the law of uninhibited growth. Given that 500 bacteria are present initially, and there are 800 after 1 hour, a growth rate of 0.470003629246 (k value) has been calculated. <span>5243 </span>will be present after 5 hours.
The law of uninhibited growth follows below formula: A = Pe^(rt) For your problem, we need to first look for the rate, given the initial and final number of bacteria.. A = Pe^(rt) 800 = (500)(e)^r(1) (800/500) = (e)^r(1) ln (8/5) = ln (e) ^ r r = 0.47 substitute the value of r to get the number of bacteria after 5 hours: A = Pe^(rt) A = (500)(e)^(0.47)(5) A = 5,242.78 or 5,243
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Step-by-step explanation:
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