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Tomtit [17]
2 years ago
6

Assume your friend just took an exam made up of 20 true/false questions. further assume that your friend had no knowledge releva

nt to the questions, and consequently just guessed the answer to each question what is the probability your friend will pass the exam?
Mathematics
1 answer:
leonid [27]2 years ago
8 0

Answer:

10/20 or 1/2 it is 50% chance that she pass and 50% that she fail

Step-by-step explanation:

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25x32/100 =8
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The distance between two cities on a map is 3.5 centimeters. The map uses a scale in which 1
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it would be 70 kilometers

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How many solutions does the following equation have? 60z+50-97z=-37z+4960z+50−97z=−37z+49
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Copy question and paste it and it will show up trust
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3 years ago
I need to fill out the blanks
zzz [600]

Answer:

1. <u>14x</u> and <u>7</u> and <u>9x</u> and <u>3</u>

2. <u>0x</u>

3. <u>4</u>

Step-by-step explanation:

Step 1: Distribute

First, multiply 7 by 2x and 1 to get 14x and 7. Then multiply -3 by 3x and 1 to get -9x and -1

Step 2: Combine like variable terms

Combine 14x, -5x, and -9x, to get 0x.

Step 3: Combine like constant terms

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6 0
3 years ago
The graph below shows the average Valentine’s Day spending between 2003 and 2012
Lapatulllka [165]
The average rate of change of a graph between two intervals is given by the difference in value of the values on the graph of the two interval divided by the difference between the two intervals.

Part A.

From the graph the average Valentine's day spending in 2005 is 98 while the average Valentine's day spending in 2007 is 120.

The average rate of change in spending between 2005 and 2007 is given by

\frac{120-98}{2007-2005} = \frac{22}{2} =\$11/year



Part B

From the graph the average Valentine's day spending in 2004 is 100 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2004 and 2010 is given by

\frac{103-100}{2010-2004} = \frac{3}{6} =\$0.5/year



Part C:

From the graph the average Valentine's day spending in 2009 is 102 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2009 and 2010 is given by

\frac{103-102}{2010-2009} = \$1/year

7 0
3 years ago
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