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beks73 [17]
2 years ago
13

Susan charges $8 per hour walking her neighbor’s dog. She charges $12 per hour babysitting. Part A: Define the variables. Part B

: Write an expression to represent the amount Susan earns dog walking and babysitting
Mathematics
1 answer:
IgorC [24]2 years ago
4 0

Answer:

Part A: The variables are the amount charged per hour dog walking and the amount charged per hour babysitting.

Part B: y=8d+12b

Step-by-step explanation:

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Seven students, A, B, C, D, E, F, and G, have been selected to be possible characters in a school play. Only 4 people will actua
nadezda [96]

if you pick G you don't have to pick D so I didn't I picked B so that I could pick E and they didn't say any rules about F so I picked G,B,E,F

8 0
2 years ago
Elysse paid for her sandwich and drink with a $10 bill and received $0.63 in change. The sandwich cost $7.75 and sales tax was $
EleoNora [17]

Given

Elysse paid for her sandwich and drink with a $10 bill and received $0.63 in change.

The sandwich cost $7.75 and sales tax was $0.47.

Find out the  cost of her drink

To proof

Let the cost of her drink be x.

As given in the question

Elysse paid for her sandwich and drink with a $10 bill and received $0.63 in change.

Elysse paid for her sandwich and drink = 10 - 0.63              

                                                                 =  $ 9.37

sandwich cost $7.75 and sales tax was $0.47

Than the equation becomes

x = 9.37 - (7.75 + 0.47)

x = 9.37 - 8.22

x = $ 1.15

The cost of the drink is $ 1.15.

Hence proved      

4 0
2 years ago
Read 2 more answers
Jordan had the following scores on her math tests last quarter: 96, 89, 79, 85, 87, 94,
Lisa [10]

Answer:

a) 5.5

b) None

Step-by-step explanation:

The given data set is {96,89,79,85,87,94,96,98}

First we must find the mean.

\bar X=\frac{96+89+79+85+87+94+96+98}{8}=\frac{724}{8}=90.5

We now find the absolute value of the distance of each value from the mean.

This is called the absolute deviation

{|96-90.5|,|89-90.5|,|79-90.5|,|85-90.5|,|87-90.5|,|94-90.5|,|96-90.5|,|98-90.5|}

{5.5,1.5,11.5,5.5,3.5,3.5,5.5,7.5}

We now find the mean of the absolute deviations

MAD=\frac{5.5+1.5+11.5+5.5+3.5+3.5+5.5+7.5}{8} =\frac{44}{8} =5.5

The least absolute deviation is 1.5. This is not within one absolute deviation.

Therefore none of the data set is closer than one mean absolute deviation away from  the mean.

5 0
2 years ago
Poor Milhouse is hopelessly in love with Lisa. Unfortunately for Milhouse, Lisa does not feel the same way. However, Milhouse re
rewona [7]

Answer:

15.6%

Step-by-step explanation:

Since each day there is a 6% chance that Lisa smiles at him then that means that each day there is a 94% chance that Lisa does not smile at him. To find the probability of Milhouse going longer than a month (30 days) without a smile from Lisa we need to multiply this percentage in decimal form for every day of the month. This can be solved easily by putting 94% to the 30th power which would be the same, but first, we need to turn it into a decimal...

94% / 100 = 0.94

0.94^{30} = 0.156

Now we can turn this decimal into a percentage by multiplying by 100

0.156 * 100 = 15.6%

Finally, we can see that the probability that Milhouse goes longer than a month without a smile from Lisa is 15.6%

6 0
2 years ago
The number of bagels sold daily for two bakeries is shown in the table. Bakery A Bakery B 53 34 52 40 50 36 48 38 53 41 47 44 55
IrinaK [193]
You are given the number of bagels sold daily for bakeries A and B and are shown in the table above. Based on the above data, it is better to describe the centers of distribution in terms of the mean than the median. This is because all of the data from bakery A and B have values that are near the average and no data sample have possible outliers. For instance,

bakery A
mean (A) = [<span>53 + 34 + 52 + 40 + 50 + 36 + 48 + 38]/8 = 43.88

Bakery B
mean (B) =  [</span>53 + 41 + 47 + 44 + 55 + 40 + 51 + 39]/8 = 46.25

Even if bakery A and B has the 38 and 39 as the lowest data respectively, they are still near the average data.
7 0
2 years ago
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