Answer:a) P(8 of the players numbers are drawn)=1.3×10^-8
b) P(7 of the players number are drrawn)=3.33×10^-c) P(at least 6 of the players number were drawn)=1.84×10^-4
Step-by-step explanation:
Players has 8 combinations of numbers from 1-40. The outcome S contains all the combinations of 8 out of 40
a) P(8 of the players numbers are drawn)= 1/40/8= 1.3×10^-8
There are one in hundred million chances that the draw numbers are precisely the chosen ones.
b) Number of ways of drawing 78 selected numbers from 1-40=8×(40-7)
8×32
P(7 of the players number are drawn)=8×32/40 =3.33×10^-6.
There are approximately 300,000 chances that 7 of the players numbers are chosen
c) P(at least 6 players numbers are drawn)= 32/2×(8/6) ways to draw.
P(at least 6 players numbers are drawn)=P(all 8 chosen are drawn)+P(7 players numbers drawn)+P(6 chosen are drawn) = 1+ 8 x32/40/8 +[8\6 ×32/2]
P(at least 6 players numbers are drawn) = 1.84×10^-4.
There are approximately 5400chances that at least6 of the numbers drawn are chosen by the player.
Answer:
B
Step-by-step explanation:
A P E X
The sides of a right triangle can always be expressed as:
h^2=x^2+y^2, where h is the length of the hypotenuse and x and y are the side lengths.
If we are to assume that "a" is a side length then 17 or "c" must be the hypotenuse so:
17^2=11^2+a^2
a^2=289-121
a^2=168
a=√168
a≈12.96 (to nearest hundredth)
Note we knew this assumption because of the answer choices, but technically, "a" COULD have been the hypotenuse without this implicit suggestion making:
a^2=17^2+11^2
a^2=410
a=√410
a≈20.25 (to the nearest hundredth)
Of course the above is not relevant to this particular question but be aware that you won't always be given answer choices to make the assumption of which side is the hypotenuse...
Solve the inequality c - 12 > -16.
Plot the solution on the number line.
