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ch4aika [34]
3 years ago
9

At Elisa's Printing Company LLC there are two kinds of printing press: Model A which can print 70 books per day and Model B whic

h can print 55 books per day. The company owns 14 total printing presses and this allows them to print 905 books per day. How many of each type of press do they have?
Mathematics
1 answer:
NeX [460]3 years ago
8 0

Answer:

Number of Model A press = 9 and Number of Model B press = 5

Step-by-step explanation:

Number of books Model A can print per day = 70

Number of books Model B can print per day = 55

Let x be the number of Model A press and y be that of Model B press.

Total number of printing presses = 14

x + y = 14  ----------------1

Total number of books printed per day = 905

70x + 55y = 905-------2

Putting y = 14 - x in equation 2, we get:

70x + 55(14 - x) = 905

15x = 135

x = 9

y = 14 - 9 = 5

Hence number of Model A press is 9 and that of Model B press is 5.

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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
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Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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3 years ago
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