Prove algebraically that the recurring decimal 0.72 =8/11
2 answers:
Answer:
see explanation
Step-by-step explanation:
We require 2 equations with the recurring decimal placed after the decimal point.
let x = 0.7272.... (1) ← multiply both sides by 100
100x = 72.7272... (2)
Subtract (1) from (2) thus eliminating the recurring decimal
99x = 72 ( divide both sides by 99 )
x = =
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P(B/A) is in the simplest form
Step-by-step explanation:
Let us explain how to solve the problem
- P(B|A) is called the "Conditional Probability" of B given A
- Conditional probability is the probability of one event occurring with some relationship to one or more other events that means event A has already happened, now what is the chance of event B
- The formula for conditional probability is P(B|A) = P(A and B)/P(A), where P(A and B) is P(A) . P(B given A occurred)
∵ There are 3 red marbles, 5 green marbles, and 2 blue
marbles in the jar
∴ The total number of the marbles in the jar = 3 + 5 + 2 = 10
∵ Two marbles are drawn from the bag, one after the other without
replacement
- That means red is already occurred when green is happened,
so it is a "Conditional Probability" of green given red
P(B/A) = P(A and B)/P(A)
∵ Event A is defined as drawing a red marble from the jar on
the first draw
∴ P(A) = P(red)
∵ P(red) =
∴ P(A) =
∵ Event B is defined as drawing a green marble on the second draw
∴ P(B) = P(green)
- The number of marbles after the red is chosen is 9
∵ P(green) =
∴ P(B) =
- Find P( A and B)
∴ P(A and B) = ( ) . (
)
∴ P(A and B) =
- Simplify the fraction by divide up and down by 15
∴ P(A and B) =
∵ P(B|A) = P(A and B)/P(A)
- Substitute the values of P(A and B) and P(A) in the rule above
∴ P(B/A) =
- Simplify the fraction
∵ ÷
=
×
=
- Divide up and down by 2
∴ P(B/A) =
P(B/A) is in the simplest form
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