What is the solution set -9.5 x - 1.5>-30

Step-by-step explanation:

How did you determine that? One way is to carry out the division. However, that's laborious, especially with larger numbers. A better method is to look at the denominator of the simplified fraction. From here on, we will only be working with simplified fractions of the form  

a

b

, where  

gcd

(

a

,

b

)

=

1

and  

b

>

1

. A well known fact is that if a prime  

p

|

b

where  

p

∉

{

2

,

5

}

, then the fraction is repeating. Think about why this is.

Here's a quick proof. If some fraction  

a

b

has a terminating decimal representation, then there exists a positive integer  

k

such that  

10

k

⋅

a

b

=

n

, where  

n

is a positive integer. Let  

p

be a prime factor of  

b

, and rewrite this as  

10

k

a

=

n

b

. Then  

p

|

10

k

a

, but since  

gcd

(

a

,

b

)

=

1

,  

p

∤

a

, so  

p

|

10

k

(ie:  

p

=

2

or  

5

).

We will now proceed to the main part of this lesson: repeating decimals. We will define the period of a decimal  

0.

d

1

d

2

d

3

d

4

…

to be the least positive integer  

n

such that for some integer  

k

,  

d

i

=

d

i

+

n

for all  

i

≥

k

. For example, the period of  

1

90

=

0.0

¯¯¯

1

is  

1

because for all  

i

≥

2

,  

d

i

=

d

i

+

1

=

1

. Similarly,  

1

7

=

0.

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

142857

has a period of  

6

because  

d

i

=

d

i

+

6

for all  

i

≥

1

. An intuitive way of thinking about period is that it is the number of digits in the repetend when written optimally. ("Written optimally" informally means that you should write  

0.0

¯¯¯

1

instead of  

0.0

¯¯¯¯¯¯

11

.)

a. the residual plot for model 1 has a random pattern and is a good fit for the data.

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