Simplify 3(8x+2y) A. 30x+y B. 24x+2y C. 24x+6y D. 30xy
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This should have been worth more points, but anyways, here are the answers. Please give thanks :)
1.)Y = 5 - 3X<span>
2.) Y= </span><span>3X - 4</span><span>
3.) Y = 7 - X
4.) Y = -5X
5.) Y = </span>X +6 <span>
6.) Y = 5 -3X
7.) Y = 2 +</span>
<span>X
8.) </span>Y =
X - 3<span>
9.) Y= 2 - [</span>tex] \frac{2}{3} [/tex]<span>
10.) Y = 1 - 5X
11.) Y = 2 - </span>
<span>X
12.) Y = -5 -2X
13.) Y = X - 6
14.) Y = </span>
<span> -2X
15.) Y = 2 +</span><span>X
16.) Y = 2 +</span><span>X
17.) Y = 1 - 5X
18.) Y = </span>[<span>tex] \frac{3}{4} [/tex]</span>X + 2
1.)Y = 5 - 3X<span>
2.) Y= </span><span>3X - 4</span><span>
3.) Y = 7 - X
4.) Y = -5X
5.) Y = </span>X +6 <span>
6.) Y = 5 -3X
7.) Y = 2 +</span>
8.) </span>Y =
9.) Y= 2 - [</span>tex] \frac{2}{3} [/tex]<span>
10.) Y = 1 - 5X
11.) Y = 2 - </span>
12.) Y = -5 -2X
13.) Y = X - 6
14.) Y = </span>
15.) Y = 2 +</span>
16.) Y = 2 +</span>
17.) Y = 1 - 5X
18.) Y = </span>[<span>tex] \frac{3}{4} [/tex]</span>X + 2
Answer with Step-by-step explanation:
We are given that two independent tosses of a fair coin.
Sample space={HH,HT,TH,TT}
We have to find that A, B and C are pairwise independent.
According to question
A={HH,HT}
B={HH,TH}
C={TT,HH}
{HH}
={HH}
={HH}
P(E)=
Using the formula
Then, we get
Total number of cases=4
Number of favorable cases to event A=2
Number of favorable cases to event B=2
Number of favorable cases to event C=2
If the two events A and B are independent then
Therefore, A and B are independent
Therefore, B and C are independent
Therefore, A and C are independent.
Hence, A, B and C are pairwise independent.