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jenyasd209 [6]
3 years ago
13

Next weekend Marnie wants to attend either carnival A or carnival B. Carnival A charges $6 for admission and an additional $1.50

per ride.
Carnival B charges $2.50 for admission and an additional $2 per ride.
a) In function notation, write A(x) to represent the total cost of attending carnival A and going on x rides.
A)In function notation,
write B(x) to represent the total cost of attending carnival B and going on x rides.
b) Determine the number of rides Marnie can go on such that the total cost of attending each carnival is the same. ​
Mathematics
1 answer:
vekshin13 years ago
4 0

Answer:

Step-by-step explanation:

carnival A : $ 6 dollar admission and $ 1.50 per ride

A(x) = 1.5x + 6

carnival B : $ 2.50 admission and $ 2 per ride

B(x) = 2x + 2.5

how many rides can Marnie go on so that the total cost of attending both is the same...

1.5x + 6 = 2x + 2.5

6 - 2.5 = 2x - 1.5x

3.5 = 0.5x

3.5 / 0.5 = x

7 = x <=== she would have to go on 7 rides for them to both cost the same

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A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is <u>n for which M = 0.04.</u>
  • There is no estimate, hence \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2

n = 600.25

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

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