5·7+(3z+0·3) = solve
1 answer:
You might be interested in
t N.
In the figure shown below
Answer:
A horizontal line segment M K intersects with line segment J L at their midpoint N.
∠J N M =(5x+2)°
∠ LN M=3( x+ 14)°
So, ∠J N M + ∠ LN M =180°[ These two angles form linear pair.Angles forming linear pair are supplementary.]
⇒5 x+ 2+ 3 (x+ 14) =180 [ By Substitution]
⇒ 5 x+2 +3 x+42°= 180°
⇒ 8 x=180°-44°
⇒8 x= 136°
⇒x= 136°÷8
⇒x=17°
So, ∠J N M =5×17 +2=87°
∠ LN M= 3×(17 +14)=3×31=93
∠J N M =∠K N L [Vertically opposite angles]
∠K N L=87°
You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.
To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.
First question:
To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is
Second question:
To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is
Third question:
To get exactly one six, it can either be the first, second or third roll.
In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.
In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus
And since there are three of these combinations, The answer is
Fourth question:
Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is
