Assume that 3 digits are selected at random from the set {1,3,5,6,7,8} { 1 , 3 , 5 , 6 , 7 , 8 } and are arranged in random orde
Mumz [18]
Sample space is 36.so the probability must be 20/36.from three digits the maximum no.of number can be made is 6.
P(being delayed on M1) = 0.5 →P(NOT being delayed on M1) = 1-0.5 = 0.5
P(being delayed on M42)=0.75→P(NOT being delayed on M42) = 1-0.75= 0.25
P(NOT being delayed on M1) AND P(NOT being delayed on M42) = 0.5x0.25
→ 0.125 = 1/8
So, the chances he will not be delayed on his journey is 1/8
Answer:
The probability that 2 certain people will serve on that committee is 11.11%.
Step-by-step explanation:
Since to make a committee 4 men are chosen out of 6 candidates, to determine what is the probability that 2 certain people will serve on that committee the following calculation must be performed:
4/6 = 2/3
1/3 x 1/3 = X
0.333 x 0.333 = X
0.1111 = X
Therefore, the probability that 2 certain people will serve on that committee is 11.11%.
<h2>7x + 19 = 208</h2><h2>11x - 89 = 208</h2><h2>--------------------------------------</h2>
<u>Step-by-step explanation:</u>
let third angle be x
60° + 60° + x = 180° (Δ sum property)
x = 60°
so, this is a equilateral triangle (all Δ equal)
<h2>-----------------------------------------</h2>
in equilateral triangle --> all sides are equal
so, therefore =>>
7x + 19 = 11x - 89
19 + 89 = 11x - 7x
108 = 4x
108 ÷ 4 = x
27 = x
<h3>x = 27</h3><h2>-----------------------------------------</h2>
so variable
1.) 7x + 19
= 7(27) + 19
= 189 + 19
= 208
2.) 11x - 89
= 11(27) - 89
= 297 - 89
= 208
