Im not entirely sure but i thinks its c
The mass of the first shipment at time t is
![m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}}](https://tex.z-dn.net/?f=m_%7B1%7D%3D100%28%20%5Cfrac%7B1%7D%7B2%7D%20%29%5E%7B%20%5Cfrac%7Bt%7D%7B10%7D%7D%20)
The mass of the second shipment at time t is
![m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }](https://tex.z-dn.net/?f=m_%7B2%7D%3D100%28%20%5Cfrac%7B1%7D%7B2%7D%20%29%5E%7B%20%5Cfrac%7Bt-3%7D%7B10%7D%20%7D)
At time t, the ratio of m₁ to m₂ is
![\frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123](https://tex.z-dn.net/?f=%20%5Cfrac%7Bm_%7B1%7D%7D%7Bm_%7B2%7D%7D%20%3D%20%5Cfrac%7B100%7D%7B100%7D.%20%5Cfrac%7B%281%2F2%29%5E%7Bt%2F10%7D%7D%7B%281%2F2%29%5E%7B%28t-3%29%2F10%7D%7D%20%5C%5C%20%3D%20%20%5Cfrac%7B%281%2F2%29%5E%7Bt%2F10%7D%7D%7B%281%2F2%29%5E%7Bt%2F10%7D%7D.%20%5Cfrac%7B1%7D%7B%281%2F2%29%5E%7B-3%2F10%7D%7D%20%5C%5C%20%3D%20%281%2F2%29%5E%7B3%2F10%7D%20%5C%5C%20%3D%200.8123)
Therefore as a percentage,
![\frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%](https://tex.z-dn.net/?f=%20%5Cfrac%7Bm_%7B1%7D%7D%7Bm_%7B2%7D%7D%20%3D100%2A0.8123%20%3D%2081.23%20%5C%25)
Answer: B. 81.2%
Answer:
The left side 45.041¯645.0416‾ is not less than the right side 42 which means that the given statement is false.
Answer: This question doesn't have an answer.
Step-by-step explanation:
That is fractional, it can not be represented as a whole number, though it can be represented as an improper or mixed fraction because it is a rational number.
629/100 or 6+29/100