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Scrat [10]
4 years ago
11

in the fall semester of 2003, the average graduate management admission test (GMAT) of the students at UTC was 500 with a standa

rd deviation of 80. in the fall of 2004, the average GMAT was 560 with a standard deviation of 84. which year's GMAT scores show a more dispersed distribution
Mathematics
1 answer:
MAVERICK [17]4 years ago
3 0

Answer:

The 2003 scores have a higher coefficient of variation, so they are more dispersed

Step-by-step explanation:

We have to find the coefficient of variation(CV) for both these tests.

CV = \frac{\sigma}{\mu}

In which \mu is the mean and \sigma is the standard deviation.

Whoever has the higher CV has the higher variation, that is, is more dispersed.

2003

\mu = 500, \sigma = 80

CV = \frac{\sigma}{\mu} = \frac{80}{500} = 0.16

2004

\mu = 560, \sigma = 84

CV = \frac{\sigma}{\mu} = \frac{84}{560} = 0.15

The 2003 scores have a higher coefficient of variation, so they are more dispersed.

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Consider the following expression and determine which statements are true. m+(5n)(9-p)-6-r^2
Fed [463]

Answer:

(B)The expression (5n)(9-p) is the product

(D)The expression 9-p has exactly two terms

Step-by-step explanation:

In the expression;

m+(5n)(9-p)-6-r^2

The coefficient of m is 1, therefore Option A is not true

The product of 5n and 9-p is (5n)(9-p), therefore Option B is true.

The expression 9-p has exactly two terms,9  and p.

Therefore, Options B and D are true.

6 0
3 years ago
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Solve the equation for y<br><br> 6y+x=8
Igoryamba

Answer:

y=4/3-1/6x

Step-by-step explanation:

1) subtract "x" from both sides:

6y=8-x

2)divide by 6 on both sides(EVERYTHING WILL BE DIVIDED BY 6)

6y/6=(8-x)/6

y=(8-x)/6

2)simplify:

y=4/3-x/6

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4 years ago
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Draw a Picture graph that
Ulleksa [173]

Got your answer.

I drew it out on a digital drawing software I have.

4 0
4 years ago
B(n)=-1(2)n-1 what's the 5th term in the sequence
Katena32 [7]
B5 = -1 x (2 ^ 5 - 1)
B5 = -1 x (2 ^ 4)
B5 = -1 x 16
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8 0
3 years ago
A park has a large circle painted in the middle of the playground area. The circle is divided into 444 equal sections, and each
balu736 [363]

Answer:

The area of each section of the circle is A=25\pi\ m^{2}

Step-by-step explanation:

we know that

Each section represent a quarter of circle

The area of a quarter of circle is equal to

A=\frac{1}{4}\pi r^{2}

we have

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substitute

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A=25\pi\ m^{2}

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