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Anastaziya [24]
3 years ago
8

PLZ HELP ASAP AREA OF A SECTOR

Mathematics
1 answer:
const2013 [10]3 years ago
4 0
The fraction of a circle represented by the shaded region is:
 S '/ S = ((8/5) * pi * r) / (2 * pi * r)
 Rewriting we have:
 S '/ S = ((8/5)) / (2)
 S '/ S = 8/10
 S '/ S = 4/5
 Then, we can make the following rule of three:
 (64/5) pi -------> 4/5
 x ----------------> 1
 Clearing x we have:
 x = ((1) / (4/5)) * ((64/5) pi)
 Rewriting:
 x = (5/4) * ((64/5) pi)
 x = (5/4) * ((64/5) pi)
 x = 50.24
 Answer:
 
The area of the complete circle is:
 
x = 50.24
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Help find the answer for problem number 4
Olin [163]
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Child ticket (c) = $2

785 tickets = $3280

a + c = 785 tickets
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c = 215 child tickets
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570 + 215 = 785 tickets
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3 0
4 years ago
If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No
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If A and B are independent, then P(A\cap B)=P(A)P(B).

a.

P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)

P(A\cup B)=0.5+0.2-0.5\cdot0.2

\boxed{P(A\cup B)=0.6}

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)

P(A^c\cap B^c)=1-0.6

\boxed{P(A^c\cap B^c)=0.4}

c. DeMorgan's law can be used here too:

P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)

P(A^c\cup B^c)=1-0.5\cdot0.2

\boxed{P(A^c\cup B^c)=0.9}

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3 years ago
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