Your answer would be 0.05
Answer:tanQ=
adjacent
opposite
=
24
7
≈0.2917≈0.29
Step-by-step explanation:
trust me bro
Answer: You can multiply the top equation by -1 to eliminate the x variable.
And the solution is (2,4/3) in case you need it.
Step-by-step explanation:
2x + 3y = 8
2x + 6y = 12
If you multiply the upper equation or down equation by one, you will be able to eliminate the x variable.
-1( 2x + 3y) = -1(8) New equation: -2x -3y = -8.
Add the new equation you got by multiplying the top equation by -1 to the bottom equation.
Add them: -2x -3y = -8
2x + 6y = 12
3y = 4
y = 4/3
You can now input the value for y into the one of the equations and solve for x.
-2x - 3(4/3) = -8
-2x -4 = -8
+4 +4
-2x = -4
x = 2
Answer:
f(2) = 2.5×2-10.5
=5-10.5
=-5.5
g(2) = 64(0.5×2)
= 64×1
=64
f(3)=2.5×3-10.5
=7.5-10.5
=-3
g(3)=64(0.5×3)
=64×1.5
=96
f(4)=2.5×4-10.5
=10-10.5
=-500
g(4)=64(0.5×4)
=64×2
=128
f(5)=2.5×5-10.5
=12.5-10.5
=2
g(5)=64(0.5×5)
=64×2.5
=150
I think it is the processs It will help you
Complete question:
Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.
Answer:
11/288
Step-by-step explanation:
We are given:
Box 1: ( 4White, ORed)
Box 2: (3White, 1Red)
Box 3: (2White, 2 Red)
We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.
Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.
Our Event, E = getting 2 red balls.
Now Box 1 is ruled out, we have:
P[E(B1)]= 0
P[E/B3)] = (2 2) / (4 2)
= 1/6
If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:
P(E/B2) = (1/4) *(1/4)
= 1/16
Therefore probability that 2 balls extracted are red, we have:
P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

= 11/288