Ask question

Ask question

All categories

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?

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%.

You might be interested in

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.

The gradient is -3/4 since it goes down 3 every 4 tiles.

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