<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>
I think it's the bar graph.
Hope this helps !
Photon
according to the question
3x-1=0
3x=1
x=⅓
so
f(x)=18x³+x-1
f(⅓)=18.(⅓)³+⅓-1
f(⅓)=18.⅓.⅓.⅓+⅓-1
f(⅓)=6.⅑+⅓-1
f(⅓)=⅔+½-1
f(⅓)=0
<h3>therefore</h3><h3> the remainder is 0</h3>
For starters, create an equation to show David's earnings. We can do this using Danielle's as a basis, which is set up as y=(# of hours)x+(bonus). This gives us y=12x+80. Now, as we need both their ys to be equal, we just set both equations equal to each other, making 15x+50=12x+80. Now, we solve for x, starting with 15x+50=12x+80, subtracting 12x from both sides to get 3x+50=80, subtracting 50 from both sides to get 3x=30, and dividing three from both sides to get x=10. To check, we just plug in our answer to both equations and see if the ys match up. With Danielle's equation, we get y=15(10)+50=150+50=200 and with David's equation, we get y=12(10)+80=120+80=200, proving that our answer is correct.
