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hjlf
4 years ago
6

The printer can print 168 papers in 21 minutes. Enter the number of paper the printer can print in 10 minutes.

Mathematics
1 answer:
kolbaska11 [484]4 years ago
4 0

Answer:

80 pages

Step-by-step explanation:

We can use ratios to solve

168 papers          x papers

------------------ = -----------------

21 minutes          10 minutes

168*10 = x * 21

1680 = 21x

Divide each side by 21

1680/21 = 21x/21

80 = x

80 pages

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If the length of the minute hand of the clock in London commonly known as "Big Ben" is 11.25' from the center of the clock to th
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Answer:

33.15 square inches.

Step-by-step explanation:

Given:

Length of the minute hand is 11.25' from the center of the clock to the tip of the hand.

<u>Question asked:</u>

What is the area swept out by the minute hand as it moves 5 minutes of time?

Solution:

<u><em>As length of the minute hand is '11.25' from the center of the clock to the tip of the hand means radius of the circle, r = 11.25 inches.</em></u>

First of all we will find area of circle swept by the minutes hand,

Area\ of\ circle=\pi r^{2}

                        =\frac{22}{7} \times11.25\times11.25\\\\ =\frac{2784.375}{7} \\ \\ =397.77\ square\ inches

When minute hand moves 60 minutes, it makes one circle of area 397.77 square inches.

By using<u> unitary method</u>, we will calculate area swept by minute hand in 5 minutes:-

In 60 minutes, area swept = 397.77 square inches

In 1 minute, area swept = \frac{397.77}{60}

In 5 minutes, area swept = \frac{397.77}{60}\times5=\frac{1988.85}{60} =33.15\ square\ inches

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8 0
4 years ago
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Answer:

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Step-by-step explanation:

The sample drawn is of size, <em>n</em> = 5046.

As the sample size is large, i.e. <em>n</em> > 30, according to the Central limit theorem the sampling distribution of sample proportion will be normally distributed with mean \hat p and standard deviation \sqrt{\frac{\hat p (1-\hat p)}{n} }.

The mean is: \hat p=0.31

The confidence level (CL) = 98%

The confidence interval for single proportion is:

CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]

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The 98% confidence interval for population proportion is:

CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]\\=[0.31-2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} },\ 0.31+2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} } ]\\=[0.31-0.0152,\ 0.31+0.0152]\\=[0.2948,0.3252]\\\approx[0.30,\ 0.33]

Thus, the 98% confidence interval [0.30, 0.33] implies that there is a 0.98 probability that the population proportion of people who refuse evacuation is between 0.30 and 0.33.  

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Answer:

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