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Misha Larkins [42]
1 year ago
14

The mean exam score for the first group of twenty examinees applying for a security job is 35.3 with a standard deviation of 3.6

.
The mean exam score for the second group of twenty examinees is 34.1 with a standard deviation of 0.5. Both distributions are close to symmetric in shape.

Use the mean and standard deviation to compare the scores of the two groups.
Mathematics
1 answer:
babymother [125]1 year ago
6 0

The scores of two groups can be compared using coefficient of variation;

Coefficient of variation (C.V.) = (Standard Deviation/ Mean) × 100%;

For Data set 1;

Standard deviation = 3.6

Mean = 35.3

C.V. = (3.6/35.3) × 100%;

       = 10.19%

For Data set 2;

Standard deviation = 0.5

Mean = 34.1

C.V. = (0.5/34.1) × 100%;

       = 1.46%

To learn more about coefficient of variation, visit: brainly.com/question/24131744

#SPJ9

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Answer:

(2,-5) and (2,1).

Step-by-step explanation:

We need to find the find all the points having an x coordinate of 2 whose distance from the point (-2,-4) is 5.

A circle with center (-2,-4) and radius 5 represents all the points whose distance from the point (-2,-4) is 5.

Standard form of a circle is

(x-h)^2+(y-k)^2=r^2

where, (h,k) is center and r is the radius.

(x-(-2))^2+(y-(-4))^2=(5)^2

(x+2)^2+(y+2)^2=25

Now, put x=2.

(2+2)^2+(y+2)^2=25

(4)^2+(y+2)^2=25

(y+2)^2=25-16

(y+2)^2=9

Taking square root on both sides.

y+2=\pm\sqrt{9}

y+2=\pm3

y+2=-3\text{ or }y+2=3

y=-5\text{ or }y=1

So, the y-coordinates of the points are -5 and 1.

Therefore, the required points are (2,-5) and (2,1).

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