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Tom [10]
3 years ago
15

A bookstore owner has 48 science fiction books and 30 mysteries.He wants the most books possible in each package, but all packag

es must contain the same number of books. How many packages can he make? How many packages of each type of book does he have?
Mathematics
2 answers:
GaryK [48]3 years ago
8 0
The bookstore owner can make six packages. There will be eight science fiction books and five mystery books in each package. This is because 6 is the highest common factor of both 48 and 30, and 6 x 8 = 48 and 6 x 5 = 30.
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likoan [24]3 years ago
4 0
<span>The bookstore owner can make six packages. There will be eight science fiction books and five mystery books in each package. This is because 6 is the highest common factor of both 48 and 30, and 6 x 8 = 48 and 6 x 5 = 30.</span>
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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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