13
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3
+
2
i
Multiply by
−
3
−
2
i
−
3
−
2
i
to make the denominator of
13
−
3
+
2
i
real.
13
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3
+
2
i
−
3
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2
i
−
3
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2
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Combine.
13
(
−
3
−
2
i
)
(
−
3
+
2
i
)
(
−
3
−
2
i
)
Expand
(
−
3
+
2
i
)
(
−
3
−
2
i
)
using the FOIL Method.
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13
(
−
3
−
2
i
)
−
3
⋅
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3
−
3
(
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2
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)
+
2
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⋅
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3
+
2
i
(
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2
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)
Simplify.
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13
(
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3
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2
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)
9
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4
i
2
Simplify each term.
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13
(
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3
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2
i
)
9
+
4
Reduce the expression by cancelling the common factors.
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−
3
−
2
i
Um I think the answer is 3
Answer:
Step-by-step explanation:
We can write two equations in the two unknowns using the given relations. Let g and b represent the costs of a round of golf and a turn in the batting cage, respectively.
5g +4b = 60 . . . . . Sylvester's expense
3g +6b = 45 . . . . . Lin's expense
Dividing the second equation by 3 gives ...
g +2b = 15 ⇒ 2b = 15 -g
Substituting into the first equation, we have ...
5g +2(2b) = 60
5g +2(15 -g) = 60 . . . . . substitute for 2b
3g = 30 . . . . . . . . . subtract 30, collect terms
g = 10 . . . . . . . divide by 3
__
2b = 15 -10 = 5 . . . . use the value of g to find b
b = 2.5 . . . . . . . . divide by 2
Mini golf costs $10 per round; batting cages cost $2.50 per turn.
Tossing a coin is a binomial experiment.
Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.
All of these trials are independent since the result of one trial does not affect the result of the next trial.
Now, for 'n' repeated trials the total number of successes is given by

where 'r' denotes the number of successful results.
In our case
and
,
Substituting the values we get,


Therefore, there are 1352078 ways to get heads if a person tosses a coin 23 times.
Answer:
15
Step-by-step explanation:
By the diagram, you can see the sum of segment AB and segment BC is segment AC. Adding the given expressions for AB and B is
. Simplifying the equation
, gives
. Substituting
in the equation for segment AC gives 15.