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BigorU [14]
3 years ago
6

I need help finding the area of 32,33,and 34

Mathematics
1 answer:
denis23 [38]3 years ago
5 0
Area is 28 for number 33.

7*8=56
56/2=28
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Answer:

0.3907

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We are given that 36​% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this​ area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly​ selected, 3 or fewer are in excellent health.

i.e. P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)

Formula :P(x=r)=^nC_r p^r q ^ {n-r}

p is the probability of success i.e. p = 0.36

q = probability of failure = 1- 0.36 = 0.64

n = 11

So, P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)

P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}

P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!}  (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}

P(x\leq 3)=0.390748

Hence  the probability that when 11 adults are randomly​ selected, 3 or fewer are in excellent health is 0.3907

5 0
3 years ago
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Arte-miy333 [17]

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start by expanding so it is written as 2÷(x+y)/(x+y)÷2.

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liubo4ka [24]

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I believe the answer is 4
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