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natita [175]
2 years ago
15

27. The attendance at a party consisted of 35 men, a number of women and some children. The number of women was one and a half t

hat of the children present. a) If there are a total of 65 participants, how many women attended the party? ​
Mathematics
1 answer:
cluponka [151]2 years ago
7 0

Answer:

<u>There are 35 Men, 12 Children, and 18 Women</u>

Step-by-step explanation:

Men(M), Women(W) and Children(C)

We learn that:

M = 35

W = 1.5C , and

M+W+C = 65

----

M+W+C = 65

M+1.5C+C = 65

35+1.5C+C = 65

2.5C = 30

C = 12

If C=12 and W=1.5C, then W=18

----

<u>There are 35 Men, 12 Children, and 18 Women</u>

(35+12+18) = 65 participants

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3 years ago
Based on information from Harper’s Index, 37% of adult Americans believe in Extraterrestrials. Out of a random sample of 100 adu
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Answer:

z(s)  is in the rejection zone , therefore we reject H₀

We have enough evidence to claim the proportion of individuals who attended college and believe in extraterrestrials is bigger than 37%

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We have a prortion test.

P₀  =  37 %         P₀  = 0,37

sample size  =  n  =  100

P sample proportion   =   P  = 47 %        P  =  0,47

confidence interval  95 %

α  =   0,05  

One tail-test  (right tail) our case is to show if sample give enough information to determine if proportion of individual who attended college is higher than the proportion found by Harper´s index.

1.-Hypothesis:

H₀      null hypothesis                        P₀  =  0,37

Hₐ  alternative hypothesis                P₀  >  0,37

2.-Confidence interval 95 %

α  =   0,05        and    z(c)  =  1.64

3.-Compute of z(s)

z(s)  =  [  P  -  P₀  ]  /√(P₀Q₀/n) ]  

z(s)  =  [ (  0,47  -  0,37  ) /  √0.37*0,63/100

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4.-Compare  z(s)   and  z(c)

z(s) > z(c)        2.08  > 1.64

5.-Decision:

z(s)  is in the rejection zone , therefore we reject H₀

We have enough evidence to claim the proportion of individuals who attended college and believe in extraterrestrials is bigger than 37%

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4 years ago
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