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2 years ago

The fifth-grade teachers are planning a field trip. There are 138 fifth-

Mathematics

2 answers:

MrMuchimi2 years ago

5 0

Answer:

6 buses

Step-by-step explanation:

138 students / 23 students per bus = 6 buses

musickatia [10]2 years ago

5 0

Answer:

they'll need 6 buses

Step-by-step explanation:

138 students divided by 23 leads for 6 buses needed

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Which is the best estimate for the mass of a desktop computer

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A is the answer 500g

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2 years ago

mr.rogers is buying candy for his class. he'll distribute the candy equally among his 25 scholars, and he wants each scholar to

stellarik [79]

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3 years ago

Is this a Percent Increase or Decrease Problem?

nydimaria [60]

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2 years ago

Simplify the following expression by combining like terms. Select the most simplified expression

Jet001 [13]

Answer:

It is D

Step-by-step explanation:

Distribute:

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8 0

2 years ago

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