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CaHeK987 [17]
3 years ago
11

A lab technician shows up to the lab at 9am on Monday and finds a culture of bacteria x containing 5328 bacteria. Each culture s

hould be labeled with the date and time the initial bacteria were placed in the culture as well as number of bacteria initially put in the culture. The technician who started this culture noted that there were 6 bacteria initially, but forgot to put the time and date. The technician watches the culture and notes that at noon there are 10,656 bacteria in the culture. When was the culture started?
Mathematics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

9 pm on a Sunday, 55 .5 days before

Step-by-step explanation:

if there were 6 bacteria initially and from the growth rate we noticed that it doubled every 3 hours

because by 9 am Monday morning there were 5328 bacteria and then by 12 noon (3 hours later), there were 10, 656 bacteria

what we need to know is how long did it take 6 bacteria to become 5328

so

if 6 bacteria = 1 hour

12 bacteria = 3 hours

24 bacteria = 6 hours

then

5328 bacteria = x hours

taking the ratio

\frac{5328}{24} =\frac{x}{6}\\ \\222=\frac{x}{6} \\\\\\222(6) = x\\\\1332=x

we need to now convert this into hours

so we divide by 24

\frac{1332}{24} =55.5

so

9 am 55.5 days before should fall on a Sunday

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3 years ago
Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2
Inessa05 [86]

Answer:

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Step-by-step explanation:

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Proof -

We know that,

Average value of f in the interval [a, b] is -

f_{ave} = \frac{1}{b - a}\int\limits^b_a {f(x)} \, dx

Now,

Here a = 0, b = \frac{3\pi }{2}, f(x) = 5sec^{2}(\frac{x}{6} )

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f_{ave} = \frac{1}{\frac{3\pi }{2}  - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx

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      = {\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]

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⇒f_{ave} = {\frac{10 }{3\pi }

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