A) the probability it is brown would be 50%; the probability it is yellow or blue would be 35%; the probability it is not green is 95%; the probability it is striped is 0%.
B) the probability of all brown would be 12.5%; the probability that the third one is the first red one drawn is 8.1%; the probability that none are yellow is 61.4%; the probability that at least one is green is 14.3%.
Explanation:
A) The probability that it is brown is the percentage of brown we have. Brown is not listed, so we subtract what we are given from 100%:
100-(15+10+20+5) = 100-(50) = 50%. The probability that one drawn is yellow or blue would be the two percentages added together: 15+20 = 35%. The probability that it is not green would be the percentage of green subtracted from 100: 100-5=95%. Since there are no striped candies listed, the probability is 0%.
B) Since we have an infinite supply of candy, we will treat these as independent events. All 3 being brown is found by taking the probability that one is brown and multiplying it 3 times:
0.5*0.5*0.5 = 0.125 = 12.5%.
To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100-10 = 90% = 0.9, for the first two, and the probability that it IS red, 10% = 0.1, for the last:
0.9*0.9*0.1 = 0.081 = 8.1%.
The probability that none are yellow is found by raising the probability that the first one is not yellow, 100-15=85%=0.85, to the third power:
0.85^3 = 0.614 = 61.4%.
The probability that at least one is green is computed by subtracting 1-(probability of no green). We first find the probability that all three are NOT green:
0.95^3 = 0.857375
1-0.857375 = 0.143 = 14.3%.
Part A:
Given
defined by
but
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
defined by
Note that in
, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular
and
Therefore, the function is a homomorphism.
Part C:
Given
, defined by
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
, defined by
but
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
, defined by
![\left([x_{12}]\right)=[x_4]](https://tex.z-dn.net/?f=%5Cleft%28%5Bx_%7B12%7D%5D%5Cright%29%3D%5Bx_4%5D)
, where
![[u_n]](https://tex.z-dn.net/?f=%5Bu_n%5D)
denotes the lass of the integer
in
.
Then, for any
![[a_{12}],[b_{12}]\in Z_{12}](https://tex.z-dn.net/?f=%5Ba_%7B12%7D%5D%2C%5Bb_%7B12%7D%5D%5Cin%20Z_%7B12%7D)
, we have
![f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%2B%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Ba%2Bb%5D_%7B12%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Ba%2Bb%5D_4%3D%5Ba%5D_4%2B%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29%2Bf%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
and
![f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Bab%5D_%7B12%7D%5Cright%29%20%5C%5C%20%5C%5C%20%3D%5Bab%5D_4%3D%5Ba%5D_4%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29f%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
Therefore, the function is a homomorphism.
On 6/25 divide 6 by 25=.24 and 2 3/4= 2.75 you have 2 and then 3/4=.75
Answer:
3 hours
Step-by-step explanation:
4.5*(2/3)
= 3 hours
The slope is -1.
because the slope formula gives you (-1-4)/(2-(-3)) which is -5/5 which also is -1
