Dependent: Ventus has a bag of chex mix and he spent his time counting and organizing the food. There are 13/31 white pieces of chex, 10/31 pretzels, and 8/31 cracker bites. he wants to see if he can grab two white chex pieces, one at a time without replacing the first attempt. He reaches his hand into the bag and gets a white chex piece and doesn't replace it. What is the probability that Ventus will grab another white chex?
independent: Cabel has a bowl of fruit and his sister won't let him take one and look at it for some strange reason/ however, she does tell him the probability of picking each fruit out of 12. 2/6 oranges, 3/12 apples, and 5/12 bananas. He must replace every fruit he picks. What is the probability of picking an orange?
(3x2 + 9x + 6) − (8x2 + 3x − 10) + (2x + 4)(3x − 7)
First distribute the (2x+4)(3x-7) to get 6x2-2x-28
After this just add the like terms for all parts of the expression.
(3x2-8x2+6x2)+(9x-3x-2x)+(6+10-28)
x2+4x-12
So you answer is x2+4x-12
Answer:
.1<u>777</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u>7</u><u> </u><u>is </u><u>repeating)</u>
First term ,a=4 , common difference =4-7=-3, n =50
sum of first 50terms= (50/2)[2×4+(50-1)(-3)]
=25×[8+49]×-3
=25×57×-3
=25× -171
= -42925
derivation of the formula for the sum of n terms
Progression, S
S=a1+a2+a3+a4+...+an
S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d] → Equation (1)
S=an+an−1+an−2+an−3+...+a1
S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d] → Equation (2)
Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)
2S=n(a1+an)
S=n/2(a1+an)
Substitute an = a1 + (n - 1)d to the above equation, we have
S=n/2{a1+[a1+(n−1)d]}
S=n/2[2a1+(n−1)d]
X ⇒ ln (2)
e^3 x - 8/e^2 x - 4
