Answer:
(a)
The probability that you stop at the fifth flip would be
(b)
The expected numbers of flips needed would be
Therefore, suppose that 
, then the expected number of flips needed would be 1/0.5 = 2.
Step-by-step explanation:
(a)
Case 1
Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be
Case 2
Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be
Therefore the probability that you stop at the fifth flip would be
(b)
The expected numbers of flips needed would be
Therefore, suppose that , then the expected number of flips needed would be 1/0.5 = 2.
3.15 addd all of them! Then divide
Answer:
x = 32
Step-by-step explanation:
x + 10 = 42, then you subtract 10 from both sides to get 32
Hi!
This question is worded a little funny, but i think in understand what it's asking. Forgive me if i'm wrong.
Place values in math are pretty self-explanatory. It's just asking you what place each number is in.
For example:
If you were to write the number 12, the 1 would be in the tens place and the 2 would be in the ones place.
When speaking about decimals, you pretty much just have to add the letters 'th' to the end of the place value, but instead of decreasing as the numbers grow smaller, you increase and the ones place doesn't exist.
So in this case:
With 249.637, 2 is in the hundreds place. 4 is in the tens place. 9 is in the ones place. 6 is in the tenths place. 7 is in the hundreths place. 3 is in the thousandths place.
Hope this helps :)
