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kotegsom [21]
2 years ago
6

It is estimated that 89% of senior citizens suffer from sleep disorders and 9% suffer from anxiety. Moreover, 5% of senior citiz

ens suffer from both sleep disorders and anxiety.
(a)Find the probability that a senior citizen suffers from anxiety, given that he or she has a sleep disorder. Round your answer to the nearest hundredth.

(b)Given that a senior citizen suffers from anxiety, what is the probability that he or she also suffers from a sleep disorder? Round your answer to the nearest hundredth.
Mathematics
1 answer:
valkas [14]2 years ago
5 0

For a;

P(a | sd) = p(a AND sd)/P(sd) = 0.06/(0.89) = 0.674

For b;

P(ds | a) = P(ds AND a)/P(a) = 0.06/0.09 =\frac{2}{3}

If my calculations are incorrect, then sorry ;/

- <u><em>❤ 7272033Alt ❤</em></u>

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Solving the whole thing
erik [133]
Let
x---------------> distance from people living to the city center

we Know that
Zone 1 covers people living within three miles of the city center
Zone 1 ------------> [x < 3 miles]

Zone 2 covers those between three and seven miles from the center
Zone 2 ------------> [ 3 <= x < = 7  miles]

Zone 3 covers those over seven miles from the center
Zone 3 ------------> [  x > 7  miles]

<span>calculate the distance between two points to find the value of x
</span>
case A) point (0,0)  point (3,4)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(4-0)² +(3-0)²]------> √[16+9]
x=√25-------------> x=5 miles

the answer Part A)
people living in (3,4)
x=5 miles -------------> covers Zone 2 [ 3 < =x <= 7  miles]


case B) point (0,0)  point (6,5)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(5-0)² +(6-0)²]------> √[25+36]
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x=√[(y2-y1)² +(x2-x1)²]----------> √[(2-0)² +(1-0)²]------> √[4+1]
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x=√[(y2-y1)² +(x2-x1)²]----------> √[(3-0)² +(0-0)²]------> √[9]
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the answer Part D)
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x=√[(y2-y1)² +(x2-x1)²]----------> √[(6-0)² +(1-0)²]------> √[36+1]
x=√37-------------> x=6.08 miles

the answer Part E)
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x=6.08 miles -------------> covers Zone 2 [ 3 < = x <= 7  miles]

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