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3 years ago

Kylie is comparing the cost of two housekeeping companies, DoItRight and CleanIt. The table describes the costs for DoItRight.

Mathematics

1 answer:

abruzzese [7]3 years ago

7 0

Answer:

First, find the cost function of DoItRight Housekeeping:

Slope(m) = cost per hour = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{44-26}{3-1.5}=\frac{18}{1.5}=12$

The y-intercept(b), representing the initial cost, can be calculated by substituting in values to the function:

$f(x)=12x+b\\26=12(1.5)+b\\b=26-18=8$

Therefore, the function for two companies are:

The function for DoItRight is $f(x)=12x+8$
The function for CleanIt is $g(x)=10x+16$

When comparing the two functions, it's shown how CleanIt has a greater y-intercept than DoltRight, meaning that CleanIt has a greater initial cost than DotRight. The y-intercept is when the graph intercepts the y-axis, therefore, the coordinates there would be (0, y-value), which, in this case for CleanIt company, will be (0, 16). While CleanIt has a greater y-intercept(initial cost), DoItRight has a greater slope, meaning they cost more per hour.

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Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.

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Step-by-step explanation:

A = {0,1,2,3}

a): R = {(0,0),(2,2),(3,3)}

R is antisymmetric, because if the (a,b)∈R, than a=b.

R is not reflexive, because (1,1) ∉ R while 1 ∈ A.

R is transitive, because if the (a,b)∈R and (b, c) ∈ R, than a=b=c and (a,c)=(a,a)∈R.

R is not portable ordering because R is not reflexive.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric, because if the (a,b)∈R and if the (b, a) ∈ R, than a=b (since (2,0) ∈ R and (0,2) ∉ R; and (2,3) ∈ R and (3,2) ∉ R )

R is reflexive, because (a,a) ∈ R of every element a ∈ A.

R is transitive , because if the (a,b)∈R and if the ( b , c )∈R . then a = b or b = c ( since there are only two element not of the form ( a , a ) and that pair does not satisfy ( a,b ) ∈ R and ( b , a ) ∈ R ), which implies ( a , c ) = ( b , c ) ∈ R or ( a , c ) = ( a , b ) ∈ R.

R is a partial ordering, because R is reflexive, antisymmetric and transitive.

c): R = {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive, because (a,a)∈R of every element a ∈ A.

R is antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R . then a = b ( since ( 1 , 2 )∈R and ( 2 , 1 ) ∉ R; ( 3 , 1 ) ∈ R and ( 1 , 3 ) ∉ R ).

R is not transitive , because ( 3 , 1 ) ∈ R and ( 1 , 2 )∈R, while ( 3 , 2 ) ∈ R.

R is not a partial ordering. because R is not transitive .

d): R = {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R is the reflexive, because ( a , a )∈R of every elements∈A.

R is the antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R, then a = b ( since ( 1 . 2 )∈R and ( 2 . 1 )∉R; similarly, all other elements not of the form (a,a) ).

R is not transitive, because ( 1 , 2 )∈R and ( 2 , 0 )∈R, while ( 1 . 0 )∉R.

R is not a partial ordering, because R is not transitive,

e): R = { ( 0 , 0 ) , ( 0, 1 ) , ( 0 , 2 ) , ( 0 , 3 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 0 ) , ( 2 , 2 ) , ( 3 , 3 ) }

R is the reflexive , because ( a , a )∈R of every element a∈A .

R is not antisymmetric, because ( 1 , 0 )∈R and ( 0 , 1 )∈R while 0 is not equal to 1.

R is not transitive, because ( 2 , 0 )∈Rand ( 0 , 3 )∈R, while ( 2 , 3 )∉R .

R is not a partial ordering, because R is not the antisymmetric and not the transitive.

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