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8_murik_8 [283]
3 years ago
6

Why is it helpful to find the mean of the data set?

Mathematics
1 answer:
HACTEHA [7]3 years ago
7 0
The mean of a set of data is the average, so when writing a conclusion or justifying a solution, instead of listing all your data you can say "the average of the data collected was _____"
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Help :( ik I asked a questions a few mins ago my brain just isn’t it today
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I can’t see your scream is crack
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2 years ago
X^2-10x = −16 solution??
Maru [420]
X^2-10x+16= 0
(x-8)(x-2)=0
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6 0
2 years ago
Read 2 more answers
Percy said that any real number for k would cause the system of equations to have no solution. Explain the error in Percy’s stat
MatroZZZ [7]

Answer:

Except k=7, any real number for k would cause the system of equations to have no solution.

Step-by-step explanation:

In general a system of equations can be represented as ax+by=c and dx+ey=f. In order this system of equations to have NO SOLUTIONS a/d=b/a≠c/f. In our example a=6, b=4, c=14, d=3, e=2 and f=k. To apply the formula above, 6/3=4/2≠14/k. Hence k≠7. It can be concluded that except k=7, any real number for k would cause the system of equations to have no solutions.

Just for information, if k=7 the system will have infinitely many solutions.

5 0
2 years ago
Read 2 more answers
What is greater 95% or 1 1/2​
Snowcat [4.5K]
1 1/2 cause your getting half not just sum
4 0
3 years ago
1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
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