lyzab19

21.04.2021 •

Mathematics

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Ответ:

dmckinnies8

19.06.2021

Mathematics

a). $P_{t}=\frac{5}{3}P_{t-1}$

b). $A_{t}=300e^{0.1277\times t}$

c). 6429 bacteria

Step-by-step explanation:

Population of the bacteria grows exponentially.

Therefore, growth of the bacteria will be represented by the formula

$P_{t}=P_{0}e^{kt}$

Where $P_{t}$ = Population of the bacteria after time t

$P_{0}$ = Initial population of the bacteria

k = growth constant

t = time taken for growth

Now we plug in the values in the formula

$P_{t}$ = 500

$P_{0}$ = 300

Time t = 4 hours

$500=300e^{4k}$

$e^{4k}=\frac{500}{300}$

Now we take the natural log (ln) on both the sides

$ln(e^{4k})=ln(\frac{500}{300})$

4k(lne) = ln(500) - ln(300)

4k = 6.2146 - 5.7038

4k = 0.5108

k = $k=\frac{0.5108}{4}=0.1277$

a). Recursive formula for the sequence formed by the bacterial growth

Since $P_{t}=(1+r)P_{t-1}$

500 = (1 + r)300

$(1+r)=\frac{5}{3}$

r = $\frac{5}{3}-1=\frac{2}{3}$

Therefore, the recursive formula will be

$P_{t}=\frac{5}{3}P_{t-1}$

b). Explicit formula for the number of the bacteria will be

$A_{t}=300e^{0.1277\times t}$

c). We have to calculate the number of bacteria after 24 hours

$A_{t}=300e^{0.1277\times 24}$

$A_{t}=300e^{3.0648}$

$A_{t}=300\times 21.43$

$A_{t}=6429$ bacteria

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