80%
Probability of pass on second attempt is 40% x 80% = 32%
80/10=8 4x8=32 40% of 80 = 32
The experimental probability for the die in question is 12/80 = 0.15
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
well, you simply must divide 32 by 80. This will give you the percentage chance she will bowl a strike on her next ball thrown.
32 is 40 percent of 80:40% of 80= 40% * 80= 40%/100% * 80= 4 * 8= 32
32 is 40% of 80.
80%
Given the way you have worded the question I take it to mean, what is the probability of drawing at least one spade?We can do this most easily by asking first, what is the probability of drawing no spades on each of the 80 times. This is 39/52. The probability of doing this 80 times is (39/52)80.Then the probability of not doing this is 1 - (39/52)80, which is quite close to one. The probability of drawing at least one spade is almost one.
To find 80 percent of a number, multiply the number by 0.8. In this instance, 32 x 0.8 = 25.6. Therefore, 80 percent of 32 is equal to 25.6.
40% of 80 is 32%..40% of 80= 40% * 80= 0.4 * 80= 32
Probability of pass on second attempt is 40% x 80% = 32%
80% of 40 is 32. 80% = 0.8 40 * 0.8 = 32
80/10=8 4x8=32 40% of 80 = 32
It is: 0.75
32/80 = 2/5