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
Answer:
1.33×10^-6
Step-by-step explanation:
5.32 x 10^-7= 5.32×10^(-5-2)= 5.32×10^-2×10^(-5)
=0.0532×10^-5
(0.0532×10^-5) (2.5 x 10^-5)
0.0532×2.5 × 10^-5
0.133×10^-5= 1.33×10^-1 ×10^-5
1.33×10^-6
Answer:
58,187.5 m²
Step-by-step explanation:
1 cm : 25 m
7 cm = 25 × 7 m
7 × 25 = 175 m
13.3 × 25 = 332.5 m
area = 332.5 m × 175 m
That would be 8 and 9.
9*8=72
While 9+8=17
I hope this helps! (:
Answer:
Step-by-step explanation:
Theoretically, there is not square root of neither 13 nor a negative number so the special symbol is used to represent the square root of a number.
