What sample size would you need in order to have a 94% confidence interval that is 8% wide? Do this calculation twice: once with
1 answer:
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Answer:
The price of each movie is $2.25
The price of each video games is $5.41
Step-by-step explanation:
Given as :
Miguel rented 5 movies and 3 video games for a total cost = $27
Next day Miguel rented 2 movies and 6 video games for a total cost = $36
Let The cost of each movie = $m
And The cost of each video game = $v
Now, According to question
5 m + 3 v = 27 .....1
2 m + 6 v = 36 ...........2
Now, solving the equation
2 × (5 m + 3 v) - (2 m + 6 v) = 2 × 27 - 36
Or, 10 m + 6 v - 2 m - 6 v = 54 - 36
Or, (10 m - 2 m) + (6 v - 6 v) = 18
Or, 8 m + 0 = 18
∴ m = =
i.e m = $2.25
So, The price of each movie = m = $2.25
Again, put the value of m into eq 2
So, 2 × 2.25 + 6 v = 36
Or, 4.5 + 6 v = 36
Or, 6 v = 36 - 4.5
Or, 6 v = $32.5
∴ v =
i.e v = $5.41
So, The price of each video games = v = $5.41
Hence, The price of each movie is $2.25 and
The price of each video games is $5.41 Answer