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Natali5045456 [20]
3 years ago
9

In a population, 93% of the people are right-handed, and 7% of the people are left-handed. In each of the last 100 days, 1 perso

n in the population has been randomly stopped on the street, For the first 93 days, the person stopped was right-handed, and for the last 7 days, the person stopped was left-handed. If the same person is allowed to be stopped on more than 1 day, what is the probability that the next person stopped will be right-handed?
Mathematics
2 answers:
BARSIC [14]3 years ago
7 0

Answer:

D or 93%

Step-by-step explanation:


stepan [7]3 years ago
6 0
93% because you completed the original set of 100% with the 93/7 day. You can practically restart the count and have again a 93% chance, as you originally had. 
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Step-by-step explanation:

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How to do it

Complete the square for
y
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12
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y
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y
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(
y
+
6
)
2
−
36
(
y
+
6
)
2
-
36
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(
y
+
6
)
2
−
36
(
y
+
6
)
2
-
36
for
y
2
+
12
y
y
2
+
12
y
in the equation
(
x
+
5
)
2
+
y
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12
y
=
3
(
x
+
5
)
2
+
y
2
+
12
y
=
3
.
(
x
+
5
)
2
+
(
y
+
6
)
2
−
36
=
3
(
x
+
5
)
2
+
(
y
+
6
)
2
-
36
=
3
Move
−
36
-
36
to the right side of the equation by adding
36
36
to both sides.
(
x
+
5
)
2
+
(
y
+
6
)
2
=
3
+
36
(
x
+
5
)
2
+
(
y
+
6
)
2
=
3
+
36
Add
3
3
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36
36
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(
x
+
5
)
2
+
(
y
+
6
)
2
=
39
(
x
+
5
)
2
+
(
y
+
6
)
2
=
39
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