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murzikaleks [220]

3 years ago

13

A tablet data provider offers different plans for data usage. The plan Raul chose has a monthly fee of $29.99 per month for 5 GB

of data with an additional cost of $10 for each gigabyte over 5.
Raul learns that his data provider is offering an unlimited data plan for $59.99 per month. Find the minimum number of gigabytes Raul could use per month to make the unlimited plan cheaper than his current plan. Explain your solution

1 answer:

jeka57 [31]3 years ago

6 0

Answer:

The minimum number is 9 GB to make the unlimited cheaper

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- A tablet data provider offers different plans for data usage

- The plan Raul chose has a monthly fee of $29.99 per month for

5 GB of data with an additional cost of $10 for each gigabyte

over 5

* <u><em>Assume that he will use </em></u><u><em>n gigabyte per month</em></u>

∵ The monthly fee is $29.99 for 5 GB

∵ The cost per GB after the first 5 GB is $10

∵ The number of GB whose cost is $10 is (n - 5)

∴ His monthly fees = 29.99 + 10 (n - 5) ⇒ current plan

- Raul learns that his data provider is offering an unlimited data

plan for $59.99 per month

∵ The fees of the unlimited is $59.99 per month

∵ We need the unlimited cheaper than current plane

- Let the current plan greater than the unlimited plan

∴ 29.99 + 10 (n - 5) > 59.99

- Subtract both sides by 29.99

∴ 10 (n - 5) > 30

- Divide both sides by 30

∴ n - 5 > 3

- Add both sides by 5

∴ n > 8

∴ The minimum number is 9 GB to make the unlimited cheaper

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We are given the weights, in the ounces, of a sample of 12 boxes below;

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Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

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$\mu$ = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

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P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

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__

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