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Ivahew [28]
3 years ago
6

Jenny is saving money to buy a game. So far she has saved $12, which is one-fourth of the total cost of the game. How much does

the game cost
Mathematics
2 answers:
7nadin3 [17]3 years ago
7 0

Answer:

1/4th of total cost = $12

so

for total cost = $12 X 4

= $ 48

the game cost is $48

vlabodo [156]3 years ago
3 0

Step-by-step explanation:

solution:

The amount of money Jenny is saving=$12

The cost of game=1÷4of 12

1÷4×12

48

:.The total price of game=$48

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A quantity with an initial value of 600 decays exponentially at a rate of
Snowcat [4.5K]

Answer:

The value of the quantity after 87 months will be of 599.64.

Step-by-step explanation:

A quantity with an initial value of 600 decays exponentially at a rate of 0.05% every 6 years.

This means that the quantity, after t periods of 6 years, is given by:

Q(t) = 600(1 - 0.0005)^{t}

What is the value of the quantity after 87 months, to the nearest hundredth?

6 years = 6*12 = 72 months

So 87 months is 87/72 = 1.2083 periods of 6 years. So we have to find Q(1.2083).

Q(t) = 600(1 - 0.0005)^{t}

Q(1.2083) = 600(1 - 0.0005)^{1.2083} = 599.64

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8 0
3 years ago
A recent taste test showed that 85% of people liked a new chicken sandwich recently added to a restaurant’s menu. The restaurant
Vesna [10]

Answer:

a) The margin of error = 0.11

b) The interval estimate = (0.74, 0.96)

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Step-by-step explanation:

The statistical data from the population and the simulation are;

The percentage of the people that liked the new chicken sandwich = 85% = The population proportion

The percentage the restaurant manager claims liked the new sandwich = 68%

The number of trials of the simulation = 100 trials

The sample size of each trial = 30

The point estimate of each trial = 0.85

The minimum sample proportion = 0.66

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a) The confidence interval = (0.66, 0.88)

By the half the range method, the margin of error is given by half the width of the confidence interval

∴ The margin of error, MOE = (0.88 - 0.66)/2 = 0.11

b) The interval estimate are the expected range of numbers for a parameters

The interval estimate = The point estimate ± MOE

∴ The interval estimate = 0.85 ± 0.11 = (0.74, 0.96)

Given that the manager's claim of 68% is not within the interval estimate, the manager's claim is not valid based on the expected values for the people that liked the new sandwich found with the point estimate and the margin of error

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i believe 1.44444444444

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Answer:

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4 0
2 years ago
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adelina 88 [10]
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