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Oksana_A [137]
2 years ago
8

Grayson charges $35 per hour plus a $35 administration fee for tax preparation. Ian charges $45 per hour plus a $15 administrati

on fee. If h represents the number of hours of tax preparation, for what number of hours does Grayson charge more than Ian?
Mathematics
2 answers:
lianna [129]2 years ago
5 0
Grayson: h hours × (35 dollars)/hour + 35 dollars = (35h + 35) dollars 
<span>Ian: h hours × (45 dollars)/hour + 15 dollars = (45h + 15) dollars </span>

<span>35h + 35 > 45h + 15 </span>
<span>20 > 10h </span>
<span>h < 2</span>
ser-zykov [4K]2 years ago
5 0

Answer:

Number of hours for which Grayson charge more than Ian is <2.

Step-by-step explanation:

Let h represents the number of hours of tax preparation.

Amount charged by Grayson = 35h+35.

Amount charged by Ian = 45h+15

It is given Grayson charge more than Ian

So,

35h+35 > 45h+15

Or 35-15 > 45h-35h

Or,  20> 10h

2> h

Or h< 2

Number of hours for which Grayson charge more than Ian is <2.

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The parallel dotplots below display the girths (belly
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Answer:

The standard deviation for the distribution of girths is

about the same for both male and female pigs.

Step-by-step explanation:

Step 1

We will interprete the dotplots

For the Male pigs

24, 26, 26, 26, 26, 26, 26, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 30, 30, 30, 30, 30, 30, 32

Range

Maximum value - Minimum value

= 32 - 24

= 8

IQR = Interquartile range = Q3 - Q1

Q1 formula = 1/4(n + 1)th

=n = 25

= 1/4(25 + 1)

= 1/4(26)

= 26/4

= 6.5

This mean the value is between the 6th and 7th value

6th value = 26

7th value = 26

Q1 = 26 + 26/2

Q1 = 26

Q3

Q3 formula = 3/4(n + 1)th

=n = 25

= 3/4(25 + 1)

= 3/4(26)

= 78/4

= 19.5

This mean the value is between the 19th and 20th value

19th value = 30

20th value = 30

Q3= 30+ 30/2

= 30

IQR = 30 - 26

= 4

Standard deviation

Mean

First we find the mean

= 24+ 26+26+ 26+ 26+ 26+26+28+ 28+ 28+ 28+28+ 28+ 28+28+ 28+28+28+30+ 30+30+ 30+30+30+32/28

= 700/25

= 28

Standard deviation Formula = √(x - mean)²/n - 1

= √[(24-28)² + (26- 28)² + (26 -28)² + (26 - 28)² +................ (32 - 28)²]/25 - 1

= √(16 + 4 + 4+ 4+ 4+4+ 4+0+ 0+ 0+ 0+0+ 0+ 0+ 0+0+0+0+ 4+ 4+4+4+4+ 4+ 16)/25 - 1

= √80/25 - 1

=√3.333333333

1.825741858

Female Pigs

21, 23, 23, 23,23,23,23,25, 25, 25, 25, 25, 25, 25, 25, 25, 25,25, 27, 27, 27, 27,27, 27, 29

Range:

Maximum value - Minimum value

= 29 - 21

8

IQR

IQR = Interquartile range = Q3 - Q1

Q1 formula = 1/4(n + 1)th

=n = 25

= 1/4(25 + 1)

= 1/4(26)

= 26/4

= 6.5

This mean the value is between the 6th and 7th value

6th value = 23

7th value = 23

Q1 = 23 + 23/2

Q1 = 23

Q3

Q3 formula = 3/4(n + 1)th

=n = 25

= 3/4(25 + 1)

= 3/4(26)

= 78/4

= 19.5

This mean the value is between the 19th and 20th value

19th value = 27

20th value = 27

Q3= 27+ 27/2

Q3 = 27

IQR = 27 - 23

= 4

First we find the Mean

=(21+23+23+23+23+23+23+25+ 25+ 25+ 25+25+ 25+ 25+ 25+25+25+25+27+27+ 27+27+27+ 27+29)/25

= 625/25

= 25

Standard deviation Formula = √(x - mean)²/n - 1

= √[(21-25)² + (23- 25)² + (23 -25)² + (23 - 25)² +................ (29 - 25)²]/25 - 1

= √(16 + 4 + 4+ 4+ 4+4+ 4+0+ 0+ 0+ 0+0+ 0+ 0+ 0+0+0+0+ 4+ 4+4+4+4+ 4+ 16)/25 - 1

= √80/25 - 1

=√3.333333333

1.825741858

Looking at the calculation above and comparing both the girths from the male and female

We can conclude that: The standard deviation for the distribution of girths is

about the same for both male and female pigs which is 1.825741858

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C I beleive the answer is C Both data sets show multiplicative relationships.

In Data Set I, y is 5.5 times x, and in Data Set II, y is 5 times x.

Step-by-step explanation:

I beleive the answer it C thank you bye bye have a nice day hope this helped

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1 year ago
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