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Novosadov [1.4K]

3 years ago

6

Each game at an arcade costs $1.50. Chuck spent no more than $12.50 at the arcade. He bought a snack for $5.25 and spent the res

t of his money on arcade games. What is the maximum number of games Chuck could have played?

1 answer:

Elodia [21]3 years ago

6 0

4 because $12.50-$5.25=$7.25 you then divide $7.25 by $1.50 which is 4.833333 Therefore the maximum number of games Chuck can play is 4

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Answer:

a) $\sigma_{\bar x} = 1.414$

b) $\sigma_{\bar x} = 1.414$

c) $\sigma_{\bar x} = 1.414$

d) $\sigma _{\bar x} = 1.343$

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

c) When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }$

$\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }$

$\sigma _{\bar x} = 1.343$

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