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Kryger [21]
3 years ago
6

At a pet store, it costs $5 to wash a cat and $7.50 to wash a dog. Last week, the store made $60 dollars from washing cats and d

ogs. If 6 cats were washed, write and solve a linear equation to find how many dogs were washed.
Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
7 0
D = number of dogs washed
c = number of cats washed

d = (60 - 5c) / 7.50
The 7.50 should be underneath the brackets 
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Is y = -x a proportional relationship
Karo-lina-s [1.5K]

Answer:

For any proportional relationship, k=yx (such as a line that passes through the origin). Find the equation of the line by solving for y in the constant of proportionality equation. This equation y=kx is another representation of a proportional relationship.

Step-by-step explanation:

5 0
3 years ago
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What is the average of 11, 13, 18 and 22?
kirill [66]

Answer:

16

Step-by-step explanation:

To find the average, add up all the number and divide by the number of terms

(11+ 13+ 18 + 22)/4

64/4

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Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
1) What is the solution of the given system?
m_a_m_a [10]

Answer:

1  x=-2.5  y = -5.5

2.  x=5  y=1

Step-by-step explanation:

1) What is the solution of the given system?

5x-y=-7

3x-y=-2

Multiply the second equation by -1

-1*(3x-y)=-1(-2)

-3x +y = 2


Now add the first equation to the modified second equation

5x-y=-7

-3x +y = 2

------------------

2x = -5

Divide each side by 2

2x/2 = -5/2

x = -2.5

Now we need to find y

-3x+y =2

-3(-2.5) +y =2

7.5 +y =2

Subtract 7.5 from each side

7.5 -7.5 +y =2-7.5

y = -5.5


2) what is the solution of the given system?

5x+7y=32

8x+6y=46

Divide the second equation by 2

8x/2+6y/2=46/2

4x+3y =23


Multiply the first equation by 4

4 (5x+7y)=32*4

20x+28y = 128


Now multiply the modified 2nd equation by -5

-5(4x+3y )=-5(23 )

-20x -15y = -115


Lets add the new equations together to eliminate x

20x+28y = 128

-20x -15y = -115

---------------------

      13y = 13

Divide each side by 13

13y/13 =13/13

y=1

Now substitute back in to find x

5x+7y=32

5x +7(1) =32

5x +7 =32

Subtract 7 from each side

5x+7-7 =32-7

5x =25

Divide by 5

5x/5 =25/5

x=5





3 0
3 years ago
A jar contains 5 blue marbles, 7 yellow marbles, and 8 green marbles. What is the probability of randomly choosing a blue marble
Leokris [45]
I want to say about 27% but I am not completely sure
7 0
3 years ago
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