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love history [14]
2 years ago
5

The 4015n printer that is used in most of the labs has a toner cartridge that cost 40.00 and each cartridge should print 10,000

pages, a ream of 500 pages for the printer costs 6.00. Just including those costs how much does it cost to print a single page
Mathematics
1 answer:
Illusion [34]2 years ago
4 0

Answer:

1 sheet costs 1.6 cents

Step-by-step explanation:

The total cost of the toner is 40 dollars which is good for 10000 pages.

The cost of 500 sheets of paper is 6.00 dollars or 600 cents

500 sheets / 600 cents = 1 sheet /x         Cross Multiply

500 sheets  x = 600 cents                        Divide by 500

x = 600/500

x = 1.2 cents.

10000 sheets cost 40 dollars. = 4000 cents

1 sheet      costs x

10000/1 = 4000/x                          Cross multiply

10000*x = 1 * 4000                        Divide by 10000

x = 4000/10000

x = 0.4 cents.

Total Cost for 1 sheet = 0.4 cents + 1.2 cents

Total Cost for 1 sheet = 1.6 cents.

 

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3 years ago
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Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
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Solve for y ; -2y - 2x =4
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4 0
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How much bigger is the Sum of first 50 even numbers than the sum of first 50 odd numbers?
Lunna [17]

Answer:

50

Step-by-step explanation:

Sum Even numbers

n = 50

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Sum = (2 + 100)*50/2

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Sum of the first 50 odd numbers

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