Answer:
a line
Step-by-step explanation:
<span>Winning Probablity = 0.2, hence Losing Probability = 0.8
Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0)
Winning once W1 is equal to L4, winning zero times is losing 5 times.
p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5);
p (W1) + p(W0) = p(L4) + p(L5)
p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5
p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>
Answer:
160
Step-by-step explanation:
Answer:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
Step-by-step explanation:
Equation I: 4x − 5y = 4
Equation II: 2x + 3y = 2
These equation can only be solved by Elimination method
Where to Eliminate x :
We Multiply Equation I by a coefficient of x in Equation II and Equation II by the coefficient of x in Equation I
Hence:
Equation I: 4x − 5y = 4 × 2
Equation II: 2x + 3y = 2 × 4
8x - 10y = 20
8x +12y = 6
Therefore, the valid reason using the given solution method to solve the system of equations shown is:
* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.
* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.
<span>5x + 20 = -5x
5x + 5x = -20
10x = -20
x = -20/10
x = -2</span>
