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inysia [295]
3 years ago
8

each hour, the number of bacteria in Dr. Nalls petri dish tripled. what percent is the population compared to the population the

hour before?
Mathematics
1 answer:
Lelu [443]3 years ago
3 0

Answer:

300%.

Step-by-step explanation:

Suppose the population is 100,  then it rises to 300 after one hour.

So the requires percent is (300/100) * 100

= 300%.

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Solve the System of Equations below using Substitution or Elimination
11Alexandr11 [23.1K]

Answer:

(0,0) or Infinitely many solutions.

Step-by-step explanation:

-−4x−4y=0

−4x−4y+4y=0+4y(Add 4y to both sides)

−4x=4y

−4x/-4= 4y/-4

(Divide both sides by -4)

x=−y

4x+4y=0

4(−y)+4y=0

0=0

7 0
2 years ago
A function which has a constant difference per interval is?
QveST [7]

Answer:

I think A or C I'm not sure tho

5 0
3 years ago
What is the solution (pls help ASAP )
Zarrin [17]

Answer:

c  the factor of the inequality is correct - took this test already

Step-by-step explanation:

4 0
3 years ago
Can any one answer this correctly?
MakcuM [25]

Answer:

A

Step-by-step explanation:

The Y-intercept is -3.

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7 0
2 years ago
Read 2 more answers
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

4 0
2 years ago
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