There could be 2 committes. 10 divided by 5 equals 2
<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:
Step-by-step explanation:
Cost (C) depends on the number of miles traveled (x).
So C is the dependent variable.
Or C(x) if they want you to be more specific.
x is your independent variable.
Miles driven do not depend on anything.
First, we have to find the z scores of $4 and $9.50.
Z₁ = ($4 - $6.50)/$2.25 = -1.11
Z₂ = ($9.50 - $6.50)/$2.25 = 1.33
Then, using a z score table, we find the probability of 1.33 and -1.11, and subtract them to determine the probability in between.
0.9082 - 0.1335 = 0.7747 or 77.47%.
Solve the following system using elimination:
{7 x + 2 y = -19 | (equation 1)
{2 y - x = 21 | (equation 2)
Add 1/7 × (equation 1) to equation 2:
{7 x + 2 y = -19 | (equation 1)
{0 x+(16 y)/7 = 128/7 | (equation 2)
Multiply equation 2 by 7/16:
{7 x + 2 y = -19 | (equation 1)
{0 x+y = 8 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{7 x+0 y = -35 | (equation 1)
{0 x+y = 8 | (equation 2)
Divide equation 1 by 7:
{x+0 y = -5 | (equation 1)
{0 x+y = 8 | (equation 2)
Collect results:
Answer: {x = -5, y = 8
