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Kisachek [45]
2 years ago
5

Sam has 1000 to distributed among two groups equally. Later ,the first part is divided among five children and second part is di

vided among two brothers. Give the expression that represent how they money distribution between two group was dispersed
Mathematics
1 answer:
melisa1 [442]2 years ago
3 0
So basically sam split the 1000 to 2 groups so we have 1000 ÷ 2
now the first group divided the 500 (since 1000÷ 2 is 500) among five children so it would be 500 ÷ 5 which equals 100 per child and the second group divided their 500 between to boys, so you would get 500÷2 which is 250
so for group one it would be: 1000÷2÷5 which equals 100 per child and the second group is 1000÷2÷2 which equals 250 for each brother
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Answer: the probability that the class length is between 50.8 and 51 min is  0.1 ≈ 10%

Step-by-step explanation:

Given data;

lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min

hence, height = 1 / ( 52.0 - 50.0) = 1 / 2

now the probability that the class length is between 50.8 and 51 min = ?

P( 50.8 < X < 51 ) = base × height

= ( 51 - 50.8) × 1/2

= 0.2 × 0.5

= 0.1 ≈ 10%  

therefore the probability that the class length is between 50.8 and 51 min is  0.1 ≈ 10%  

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2 years ago
Michelle can fold 4 baskets of clothes in 54 minutes, while Ruby can fold 4 baskets of clothes in 108 minutes. How long will it
padilas [110]
Michelle can fold 4/54 baskets per minute = 8/108 baskets per minute.
Ruby can fold 4/108 baskets per minute.

Each minute Michelle and Ruby work together, they can fold
   8/108 +4/108 = 12/108 = 1/9
of a basket of clothes. For 8 baskets of clothes, it will take them
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3 0
3 years ago
Please help me c:
netineya [11]

Oscar played games vs number of points he scored is, C) positive, linear association.

Step-by-step explanation:

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3 years ago
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ser-zykov [4K]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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nekit [7.7K]

Answer:

option f is right

Step-by-step explanation:

Given that data is collected to perform the following hypothesis test.

H_0: \mu = 5.5\\H_a: \mu >5.5

(right tailed test)

Sample mean = 5.4

p value = 0.1034

when p value = 0.1034 we normally accept null hypothesis.  i.e chances of null hypothesis true is  the probability of obtaining test results at least as extreme as the results actually observed during the test, assuming that the null hypothesis is correct

f) If the mean µ does not differ significantly from 5.5 (that is, if the null hypothesis is true), then the probability of obtaining a sample mean y as far or farther from 5.5 than 5.4 is .1034.

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3 years ago
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