B= 20°
and c = 160°
b is equal to the given angle, b = 20°
c is 180 - 20 = 160°
Answer: ![\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
We know that probability for any event = ![\dfrac{\text{Number of favorable outcomes}}{\text{Total outcomes}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7BNumber%20of%20favorable%20outcomes%7D%7D%7B%5Ctext%7BTotal%20outcomes%7D%7D)
Given : Charlotte has 6 cherry candies, 3 grape candies, and 3 lime candies.
I..e Total pieces of candies she has = 6+3+3= 12
Now , If Charlotte randomly pulls one piece of candy out of the bag, what is the probability that it will be cherry is given by :-
![\text{P(cherry)}=\dfrac{\text{Number of cherries}}{\text{Total candies}}\\\\=\dfrac{6}{12}\\\\=\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Ctext%7BP%28cherry%29%7D%3D%5Cdfrac%7B%5Ctext%7BNumber%20of%20cherries%7D%7D%7B%5Ctext%7BTotal%20candies%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B6%7D%7B12%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D)
Hence, the probability that it will be cherry is
.
Answer:
The answer is "26179.4".
Step-by-step explanation:
Assume year 2000 as t, that is t =0.
Formula:
![A= A_0e^{rt}](https://tex.z-dn.net/?f=A%3D%20A_0e%5E%7Brt%7D)
Where,
![A_0 = \ initial \ pop \\\\r= \ rate \ in \ decimal \\\\t= \ time \ in \ year](https://tex.z-dn.net/?f=A_0%20%3D%20%5C%20initial%20%5C%20pop%20%5C%5C%5C%5Cr%3D%20%5C%20rate%20%5C%20in%20%5C%20decimal%20%5C%5C%5C%5Ct%3D%20%5C%20time%20%5C%20in%20%5C%20year)
for doubling time,
![r = \frac{log (2)}{t} \\](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Blog%20%282%29%7D%7Bt%7D%20%5C%5C)
![r = \frac{\log (2)}{ 40} \\\\r= \frac{0.301}{40}\\\\r= 0.007](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5Clog%20%282%29%7D%7B%2040%7D%20%5C%5C%5C%5Cr%3D%20%5Cfrac%7B0.301%7D%7B40%7D%5C%5C%5C%5Cr%3D%200.007)
Given value:
![A = A_0e^{rt} \\\\](https://tex.z-dn.net/?f=A%20%3D%20A_0e%5E%7Brt%7D%20%5C%5C%5C%5C)
![A_0 = 13000](https://tex.z-dn.net/?f=A_0%20%3D%2013000)
![t= 40 \ years](https://tex.z-dn.net/?f=t%3D%2040%20%5C%20years)
when year is 2000, t=0 so, year is 2100 year as t = 100.
![A = 13000 \times e^{et}\\\\A = 13000 \times e^{e \times t}\\\\A = 13000 \times e^{0.007 \times 100}\\\\A = 13000 \times e^{0.7}\\\\A= 13000\times 2.0138\\\\A = 26179.4](https://tex.z-dn.net/?f=A%20%3D%2013000%20%5Ctimes%20e%5E%7Bet%7D%5C%5C%5C%5CA%20%3D%2013000%20%5Ctimes%20e%5E%7Be%20%5Ctimes%20t%7D%5C%5C%5C%5CA%20%3D%2013000%20%5Ctimes%20e%5E%7B0.007%20%5Ctimes%20100%7D%5C%5C%5C%5CA%20%3D%2013000%20%5Ctimes%20e%5E%7B0.7%7D%5C%5C%5C%5CA%3D%2013000%5Ctimes%202.0138%5C%5C%5C%5CA%20%3D%2026179.4)
Step-by-step explanation:
Determine whether a number is a solution to an equation.
Substitute the number for the variable in the equation.
Simplify the expressions on both sides of the equation.
Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.
Hoped that helped:P