The length of pencil A is 5 cm
<em><u>Solution:</u></em>
Let the length of pencil A be "x"
Let the length of pencil B be "y"
Let the length of pencil C be "z"
<em><u>The total length of pencils A, B and C is 29 cm</u></em>
Therefore,
length of pencil A + length of pencil B + length of pencil C = 29
x + y + z = 29 ------------ eqn 1
<em><u>Pencil A is 11 cm shorter then pencil B</u></em>
x = y - 11 ------- eqn 2
<em><u>Pencil B is twice as long a pencil C</u></em>
y = 2z
------ eqn 3
<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

<em><u>Substitute y = 16 in eqn 2</u></em>
x = 16 - 11
x = 5
Thus length of pencil A is 5 cm
Answer:
a) No
b) 42%
c) 8%
d) X 0 1 2
P(X) 42% 50% 8%
e) 0.62
Step-by-step explanation:
a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.
b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6
P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7
P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
c) P(win first game) = 0.4
P(win second game) = 0.2
P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%
d) X 0 1 2
P(X) 42% 50% 8%
P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%
e) The expected value 
f) Variance 
Standard deviation 
3/5x-5
I believe this is the expression that is described.
What is the length of BD given that figures ABCD is a rectangle and AC=26