A 125-page document is being printed by five printers. Each page will be printedexactly once.(a) Suppose that there are no restrictions on how many pages a printer canprint. How many ways are there for the 125 pages to be assigned to thefive printers?

Using the Fundamental Counting Theorem, it is found that there are ways for the 125 pages to be assigned to the five printers.

Fundamental counting theorem:

States that if there are n things, each with  ways to be done, each thing independent of the other, the number of ways they can be done is:

In this problem:

5 is multiplied by itself 125 times, hence:

There are ways for the 125 pages to be assigned to the five printers.

A similar problem is given at link

5^125 ways

Step-by-step explanation:

We proceed as follows

Let us label the five printers as A,B,C,D and E

Each printer can print any number of pages since there are no restrictions

Therefore for every page to be printed , there are 5 printer options available

Page 1 = 5 options

Page 2 = 5 options and so on till page 125

So total number of ways to assign 125 pages to five printer = 5*5*5**5 (125times)

So total ways = 5^125 ways

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.