D.d=24%
d.r=29%
d.t=53%
r.d=24%
r.r=24%
r.t=47%
t.d=47%
t.r=53%
t.t=100%
Just take the numbers and divide them over the total amount which in this caase is 51.
The side adjacent to ∠W is side WU.
Answer:
The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.
Step-by-step explanation:
To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.
Both heads and tails have an individual probability p=0.5.
Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.
The probability that k heads are in the sample is:

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

For the last five tosses, the probability that are exactly 4 heads is:

Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:

Answer: A and D
Step-by-step explanation:because my teacher told me