The required probability of the coin landing tails up at least two times is 15/16.
Given that,
A fair coin is flipped seven times. what is the probability of the coin landing tails up at least two times is to be determined.
<h3>What is probability?</h3>
Probability can be defined as the ratio of favorable outcomes to the total number of events.
Here,
In the given question,
let's approach inverse operation,
The probability of all tails = 1 / 2^7 because there is only one way to flip these coins and get no heads.
The probability of getting 1 head = 7 /2^7
Adding both the probability = 8 / 2^7
Probability of the coin landing tails up at least two times = 1 - 8/2^7
= 1 - 8 / 128
= 120 / 128
= 15 / 16
Thus, the required probability of the coin landing tails up at least two times is 15/16.
Learn more about probability here:
brainly.com/question/14290572
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Answer:
Coordinates: (0,0) ; (1,5) ; (2,10)
Step-by-step explanation:
x | y
0 0
1 5
2 10
Coordinates: (0,0) ; (1,5) ; (2,10)
This should be a straight line that is going diagonally with a positive slope.
Answer:
angle B: 39
AC: 7.288
AB: 11.58
(note: you cut off the part about rounding so make sure that it's rounded correctly before you put in your answer)
Step-by-step explanation:
To solve this we will use SOH, CAH, TOA
we have the angle and the one opposite to it which means we can use either SOH or TOA
let's use TOA
tan(51)=(9/x)
x= 7.288
We can now use pahtagaryous theroem to solve for the hyptonouse
we have
9²+7.288²=C²
C=11.58
Finally, to find angle B we will recall that the angles of a triangle must add to 180.
51+90+a=180
a=39

so, you'd end up with a pyramid with a volume "one sixteenth" of the original then
now, the original had a volume of 1536, the quarterized version will then just be one sixteenth of that, or 1536 * 1/36