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lyudmila [28]
3 years ago
6

The manager of an amusement park recorded the number of visitors to the park. There were 800 visitors on Saturday, and 1,100 vis

itors on Sunday.
What is the percentage change from the number of park visitors on Saturday to the number of park visitors on Sunday?
Mathematics
1 answer:
kirill [66]3 years ago
5 0

The percentage change from the number of park visitors on Saturday

to the number of park visitors on Sunday is 37.5%

Step-by-step explanation:

The percentage of change from A to B is:

The percentage of change = \frac{B-A}{A} × 100%

The manager of an amusement park recorded the number of visitors

to the park

  • There were 800 visitors on Saturday
  • There were 1,100 visitors on Sunday

We need to find the percentage change from the number of park

visitors on Saturday to the number of park visitors on Sunday

∵ The number of park visitors on Saturday = 800

∵ The number of park visitors on Sunday = 1,100

- Use the rule of the percentage change above

∴ The percentage change from the number of park visitors on

  Saturday to the number of park visitors on Sunday =

   \frac{1,100-800}{800} × 100%

∴ The percentage of change = 37.5%

The percentage change from the number of park visitors on Saturday

to the number of park visitors on Sunday is 37.5%

Learn more:

You can learn more about percentage in brainly.com/question/12501490

#LearnwithBrainly

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Answer:

a) No

b) 42%

c) 8%

d) X               0                 1                2

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e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

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P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

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P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

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