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Setler79 [48]
3 years ago
13

The average mark of candidates in an aptitude test was 128.5 with a standard deviation of 8.2.Three scores extracted from the te

st are 148,102,152.What is the average of the extracted scores that are outliers
Mathematics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

102

Step-by-step explanation:

We have the mean (m) 128.5 and the standard deviation (sd) 8.2, we must calculate the value of z for each one and determine whether or not it is an outlier:

z = (x - m) / sd

In the first case x = 148:

z = (148 - 128.5) /8.2

z = 2.37

In the second case x = 102:

z = (102 - 128.5) /8.2

z = -3.23

In the first case x = 152:

z = (152 - 128.5) /8.2

z = 2.86

The value of this is usually between -3 and 3, therefore when x is 102 it goes outside the range of the value of z, which means that this is the outlier.

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Please answer ASAP!​
Scorpion4ik [409]

Multiply the two brackets together

(2x-1)(x-2)

2x(x)(2x*-2)(-1*x)(-1*-2)

2x^2-4x-x+2

2x^2-5x+2

Answer is 2x^2-5x+2 - J

8 0
3 years ago
The distance from Rodney's house to the park is 100 meters. If he walks to the park and back to his house, how far will he have
konstantin123 [22]

Answer:

c. 20,000cm

Step-by-step explanation:

from is house to the park is 100m

from the park back to his house is another 100m

100m+100m=200m

1m = 100cm

200m = xcm

cross multiply

xcm = 200×100

xcm= 20,000cm

3 0
3 years ago
The graph below represents which system of inequalities?
attashe74 [19]

Answer:

y ≤ 2x -2 Option C.

Step-by-step explanation:

Equation of the dotted line which passes through (3,0) and (0,8) will be

y = mx+C

m = \frac{y-y'}{x-x'}=\frac{3-0}{0-3} = -1

and C = 3

So equation will be y = x + 3

But shaded blue part is above the dotted line so inequality will be y> -x +3

Equation of the second solid line which passes through ( 0,-2) and (1,0) will be

y = mx + C

m = \frac{y-y'}{x-x'} = \frac{0+2}{1-0} = 2

and C = -2

So equation of the line will be y = 2x - 2

But shaded yellow part is below the solid line.

So inequality will be

y ≤ 2x -2

Option C. is the answer.

5 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
Find the area of the shaded region.
Ghella [55]
This is a rhombus and in any rhombus, the diagonals intersects in the middle and they are perpendicular:
So all 4 triangles are right triangles and the sides are the hypotenuses.
1st) Calculate the sides: hypotenuse² = 3² + 4² = 25, and hypotenuse = 5

The area of each right triangle is (4 x 3)/2 = 6 units²
And the area of the 4 right triangles = 4 x 6 = 24 init²

4 0
3 years ago
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