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3 years ago

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A pile of newspaper was 17 3/4 inches high. Each consecutive week for the next 5 weeks the height of pile increase by 8 7/12 inc

hes. What is the height in inches, of the pile after 3 weeks?

1 answer:

Murljashka [212]3 years ago

7 0

Answer: the height in inches, of the pile after 3 weeks is 34 11/12 inches

Step-by-step explanation:

Each consecutive week for the next 5 weeks the height of pile increase by 8 7/12 inches. Converting 8 7/12 inches to improper fraction, it becomes 103/12 inches. The height is increasing in an arithmetic progression. The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = 17 3/4= 71/4 inches

d = 103/12 inches

n = 3 weeks

the height in inches, of the pile after 3 weeks, T3. Therefore,

T3 = 71/4 + (3 - 1)103/12

T3 = 71/4 + 2 × 103/12 = 71/4 + 103/6

T3 = 419/12 inches = 34 11/12 inches

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Answer:

Kameryn will have more words typed than Joe when the number of minutes exceeds 34.

Step-by-step explanation:

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y ----> the total words typed

we know that

<em>Kameryn</em>

$y=75x$ -----> equation A

<em>Joe</em>

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Solve the system of equations by substitution

Substitute equation A in equation B and solve for x

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That means

For x=34 minutes

The amount of words written by Kameryn and Joe are the same.

therefore

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3 years ago

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Answer:

We need a sample of size at least 13.

Step-by-step explanation:

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

In which

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

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

$\pi = \frac{0.438 + 0.642}{2} = 0.54$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

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

$0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}$

$0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}$

$\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}$

$n = 12.21$

Rounding up

We need a sample of size at least 13.

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3 years ago