At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
I hope this helps you
✔cos^2A+sin^2A=1
✔1-cos^2A=sin^2A
✔cos2A=cos^2A-sin^2A
✔sin2A=2.sinA.cosA
secA=1/cosA
tgA=sinA/cosA
sin^2A/1/cos^2A-sin^2A/cos^2A
sin^2A.cos^2A/cos2A
2.sin^2A.cos^2A/cos2A
sin2A.2.sin2A/cos2A
tg2A.2.sin2A
Y-3 is the correct answer
Answer:
1/2
Step-by-step explanation:
9/10-1/3=0.56666666666666666...
I’m pretty sure the answer would be 4 because if you put 48 as a decimal 0.48 moving it to places to the left 4 is in the tenths place
