There are 24 hours in a day. 25% is 25 / 100
% can always be replaced by divide by 100.
25/100 * 24 = 25*24/100 = 600 / 100 = 6
Therefore someone spends 6 hours a day in school.
10% of 1000 is 100. So 3% must be 30. $1000 + 30 = 1030. However, she adds the $50 in the account as well. So 1030 + 50 = $1080
Can you please attach the enlargement? Thanks.
Answer:
0.002 mg of sample will be left after 40 days.
Step-by-step explanation:
Half life of the iodine-131 =
= 8 days
Initial amount of iodine-131 =
= 64 milligrams = 0.064 g
Amount of iodine-131 left after the time of 40 days = N
Decay constant =
= ?

![\log[N]=\log[N_o]-\frac{\lambda \times t}{2.303}](https://tex.z-dn.net/?f=%5Clog%5BN%5D%3D%5Clog%5BN_o%5D-%5Cfrac%7B%5Clambda%20%5Ctimes%20t%7D%7B2.303%7D)
![\log[N]=\log[0.064 g]-\frac{0.0866 day^{-1}\times 40 days}{2.303}](https://tex.z-dn.net/?f=%5Clog%5BN%5D%3D%5Clog%5B0.064%20g%5D-%5Cfrac%7B0.0866%20day%5E%7B-1%7D%5Ctimes%2040%20days%7D%7B2.303%7D)
![N=Antilog [-2.6979 g]](https://tex.z-dn.net/?f=N%3DAntilog%20%5B-2.6979%20g%5D%20)
N = 0.002004 g = 0.002 mg
0.002 mg of sample will be left after 40 days.
The answer is c because (-1/5)^3=-125 and (-1/5)^3=-1/125 so c is the reciprocal of the original problem