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REY [17]
3 years ago
15

A car wash attendant counted the number of cars washed and the total amount of money earned. The

Mathematics
1 answer:
Gemiola [76]3 years ago
8 0

Answer:

<em>total</em><em>=</em><em>n÷m</em><em> </em><em>$</em><em>1</em><em>6</em><em>5</em><em>÷</em><em>1</em><em>5</em><em>,</em><em>$</em><em>2</em><em>3</em><em>1</em><em>÷</em><em>2</em><em>1</em><em> </em><em>and</em><em> </em><em>$</em><em>2</em><em>7</em><em>5</em><em>÷</em><em>2</em><em>5</em>

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The Wilson family is going to an amusement park for the day. The park charges $15 per vehicle for parking. The cost of each admi
Vladimir79 [104]

Answer:

D.

Step-by-step explanation:

5 0
3 years ago
Help me please !!!!!
lys-0071 [83]

Answer:

The answer is $28,620.

Step-by-step explanation:

$53,000 * 2.7% = 1431

$1,431 * 20 = $28,620

Hope this helped you!!!

4 0
2 years ago
A waitress works 1.75 hours less in the afternoon than in the evening. If she works 5      1/8
gayaneshka [121]
Let x be the number of hours that the waitress works in the evening.

We know for our problem that she works 5 \frac{1}{8} in the afternoon and that she works <span>1.75 hours less in the afternoon than in the evening, so:
</span>x=5 \frac{1}{8} -1.75
<span>Since </span>\frac{1}{8} =0.125, we can rewrite our expression:
x=5.125-1.75
x=3.375

We can conclude that she works 3.375 hours in the evening, or expressed as a mixed fraction: 3 \frac{3}{8} hours.
3 0
3 years ago
What would a 64 ring size be in the US THANKSSSS​
Mazyrski [523]

Answer:

10.75 would be a 64 ring size in the US

5 0
3 years ago
For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)
vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

  • The relation being reflexive (∀a∈R, a∼a)
  • The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
  • The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0
3 years ago
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