From point A, draw a line segment at an angle to the given line, and about the same length. The exact length is not important. Set the compasses on A, and set its width to a bit less than one fifth of the length of the new line. Step the compasses along the line, marking off 5 arcs. Label the last one C. With the compasses' width set to CB, draw an arc from A just below it. With the compasses' width set to AC, draw an arc from B crossing the one drawn in step 4. This intersection is point D. Draw a line from D to B. Using the same compasses' width as used to step along AC, step the compasses from D along DB making 4 new arcs across the line. Draw lines between the corresponding points along AC and DB. Done. The lines divide the given line segment AB in to 5 congruent parts.
Answer:
-2 + × = y
Step-by-step explanation:
I think this is the answer because it's asking for an expression which this would be one.
T(1) = 3, t(n) = -2t(n-1) + 1
So t(2) = -2(t(1)) + 1 = -2(3) + 1 = -5
t(3) = -2(t(2)) +1 = -2(-5) +1 = 11
t(4) = -2(t(3)) +1 = -2(11) +1 = -21
t(5) = -2(t(4)) +1 = -2(-21) +1 = 43
You may go the super easy way and choose 1 as the whole number and 0.28 as the decimal number. Then you'd obviously have 
Otherwise, to be a bit more creative, you can observe that
And since 1/25=0.04, you can also choose 7 as the integer, and 0.04 as the decimal to get
Answer:
0.625
Step-by-step explanation:
Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.
Since A1 and A2 are mutually exclusive and exhaustive, we can say
b) P(B) = P(A1B)+P(A2B)
Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads.
c) Using Bayes theorem
conditional probability that it wasthe biased coin
=
d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin=
