We know that , in a circle radius perpendicular to chord will bisect the chord.
OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13
Should be the first option.
Since ABP is congruent to CBP in the 4th step, AP must be equal to CP
It would be: -2/3 * -9/10 = 18/30 = 3/5
So, option B is your answer.
Hope this helps!
Answer:
Step-by-step explanation:
We want to simplify:
This is the same as:
Cancel out the common factors to get;
This is the same as
Answer:
2/7
then 2/14
Step-by-step explanation:
Let P(H)=p be the probability of one head. In many scenarios, this probability is assumed to be p=12 for an unbiased coin. In this instance, P(H)=3P(T) so that p=3(1−p)⟹4p=3 or p=34.
You are interested in the event that out of three coin tosses, at least 2 of them are Heads, or equivalently, at most one of them is tails. So you are interested in finding the likelihood of zero tails, or one tails.
The probability of zero tails would be the case where you only received heads. Since each coin toss is independent, you can multiply these three tosses together: P(H)P(H)P(H)=p3 or in your case, (34)3=2764.
Now we must consider the case where one of your coin flips is a tails. Since you have three flips, you have three independent opportunities for tails. The likelihood of two heads and one tails is 3(p2)(1−p). The reason for the 3 coefficient is the fact that there are three possible events which include two heads and one tails: HHT,HTH,THH. In your case (where the coin is 3 times more likely to have heads): 3(34)2(14)=2764.
Adding those events together you get p3+3(p2)(1−p)=5464. Note that the 3 coefficient
