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Lisa [10]
3 years ago
14

1. The range of guest on New Year's Eve was 15>

Mathematics
2 answers:
leonid [27]3 years ago
7 0

Answer:

1.False

2. True

3.) True

4. False

5.) False

6.) True

Step-by-step explanation:

1.)

Range = maximum - minimum

New year's eve :

Maximum = 22 ; minimum = 3

Range = 22 - 3 = 19

2.)

Median for Thanksgiving = 17

50% up to 17 ;hence the remaining 50% is also 17 and above

4.)

IQR for Thanksgiving

Q3 - Q1

23 - 12 = 11

5.)

The median number of people at new year's eve for dinner was 7 (line in between the box)

6.)

3/4th of new year eve's dinner = Q3 = 15

1/4 = Q1 ; have up to 5 people

Therefore ; the remaining (1 - 1/4) = 3/4 have 5 and above

Marizza181 [45]3 years ago
3 0

Answer:

1.False

2. True

3.) True

4. False

5.) False

6.) True

Step-by-step explanation:

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Answer:

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Step-by-step explanation

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Thoughtfully split the expression at hand into groups, each group having two terms :

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Group 2:  2x3-5x2

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Pull out from each group separately :

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Bad news !! Factoring by pulling out fails :

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If 2^(x+3) - 2x = k(2^x) what is the value of k
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Step-by-step explanation:

Use:

a^n\cdot a^m=a^{n+m}

2^{x+3}-2^x=2^x\cdot2^3-2^x=8\cdot2^x-2^x=(8-1)\cdot2^x=7\cdot2^x\\\\2^{x+3}-2^x=k(2^x)\\\\7\cdot2^x=k(2^x)\Rightarrow k=7

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