what are the sentences? but 2/4, could also be 1/2 or a half
Area=length*width
(32+48)=(16)*(width)
80=(16)(width)
width=(80/16)
5=width
Answer:

Step-by-step explanation:
we are given half-life of PO-210 and the initial mass
we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>
in order to solve so we can consider the half-life formula given by

where:
- f(t) is the remaining quantity of a substance after time t has elapsed.
- a is the initial quantity of this substance.
- T is the half-life
since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5
thus substitute:

reduce fraction:

By using calculator we acquire:

hence, the remaining sample after 420 days is 0.625 kg
Answer:
The standard deviation of the age distribution is 6.2899 years.
Step-by-step explanation:
The formula to compute the standard deviation is:

The data provided is:
X = {19, 19, 21, 25, 25, 28, 29, 30, 31, 32, 40}
Compute the mean of the data as follows:

![=\frac{1}{11}\times [19+19+21+...+40]\\\\=\frac{299}{11}\\\\=27.182](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B11%7D%5Ctimes%20%5B19%2B19%2B21%2B...%2B40%5D%5C%5C%5C%5C%3D%5Cfrac%7B299%7D%7B11%7D%5C%5C%5C%5C%3D27.182)
Compute the standard deviation as follows:

![=\sqrt{\frac{1}{11-1}\times [(19-27.182)^{2}+(19-27.182)^{2}+...+(40-27.182)^{2}]}}\\\\=\sqrt{\frac{395.6364}{10}}\\\\=6.28996\\\\\approx 6.2899](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B11-1%7D%5Ctimes%20%5B%2819-27.182%29%5E%7B2%7D%2B%2819-27.182%29%5E%7B2%7D%2B...%2B%2840-27.182%29%5E%7B2%7D%5D%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B%5Cfrac%7B395.6364%7D%7B10%7D%7D%5C%5C%5C%5C%3D6.28996%5C%5C%5C%5C%5Capprox%206.2899)
Thus, the standard deviation of the age distribution is 6.2899 years.