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ser-zykov [4K]
3 years ago
15

create a list of 6 values where the mean,median,and mode are 45,and only two of thw values are the same

Mathematics
1 answer:
rodikova [14]3 years ago
8 0
Here mimiqueen
43, 44 45 45 46 47
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Write the correct inequality for each graph.
Arada [10]

1.) x>2

2.)x≤3

3.)x<4

4.)x≥5

Hope this helps:)

7 0
3 years ago
8. Solve the equation<br>x² - 1 = 48​
crimeas [40]

Answer:

x^2-1=48

x^2=48+1

x^2=49

x=\sqrt{49}

x=7

Step-by-step explanation:

5 0
3 years ago
Veronikas five test scores are 59 80 95 and 93 if the outlier of 59 is removed what is the mean absolute deviation of the remain
Anit [1.1K]

Complete Question

Veronikas four test scores are 59, 80, 95, 88 and 93 if the outlier of 59 is removed what is the mean absolute deviation of the remaining four test scores?

Answer:

5

Step-by-step explanation:

We have the four test scores

80, 95, 88 and 93

Step 1

We find the mean of the 4 test scores

= 80 + 95 + 88 + 93/4

= 356 / 4

89

Step 2

The formula for Mean Absolute Deviation =

Summation( x - Mean)/n

Hence,

|(80 - 89 )+( 95 - 89) + (88 - 89) + (93 - 89)|/5

= 9 + 6 +1 + 4/4

= 20/4

= 5

8 0
3 years ago
Management selection. A corporation plans to fill 2 different positions for​ vice-president, Upper V 1 and Upper V 2​, from
Mandarinka [93]

Answer:

45 different ways to fill the positions.

Step-by-step explanation:

Consider the provided information.

Management selection. A corporation plans to fill 2 different positions for vice-president, V₁ and Upper V₂, from administrative officers in 2 of its manufacturing plants. Plant A has 9 officers and plant B has 5.

We need to find the probability that 2 positions be filled if the Upper V₁ position is to be filled from plant A and the Upper V₂, position from plant B?

To fill the post of V₁ from plant A, which has 9 officers, we have 9 ways.

To fill the post of V₂ from plant B, which has 5 officers, we have 5 ways.

Therefore, the total number of ways are: 9\times 5=45

Hence, 45 different ways to fill the positions.

8 0
4 years ago
1. The midpoint of the segment joining points (a, b) and ( j, k) is: a. (j-a,k-b) b. ((j-a)/2,(k-b)/2) c. (j+a,k+b) d. ((j+a)/2,
harkovskaia [24]
1. The midpoint of the segment joining points (a, b) and ( j, k) is ((j+a)/2,(k+b)/2)

2. Let the coordinate of H be (a, b)
T(0, 4) = ((a + 0)/2, (b + 2)/2)
(a + 0)/2 = 0 => a + 0 = 0 => a = 0
(b + 2)/2 = 4 => b + 2 = (2 x 4) = 8 => b = 8 - 2 = 6
Therefore, the cordinate of H is (0, 6)

3. Point (-4, 3) lies in Quadrant II

4. Point (6, 0) lies on the x-axis

5. Any line with no slope is parallel to the y-axis

7. a is the value of the x-coordinate.
5a + 3 = 8
5a = 8 - 3 = 5
a = 5/5 = 1
a = 1

8. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => a = -5 and b = 7.
Therefore, its center point is (-5, 7)

9. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => r = 6.
Therefore, its radius is 6

10. If the equation of a circle is (x - 2)^2 + (y - 6)^2 = 4, the center is point (2, 6).
True

7 0
4 years ago
Read 2 more answers
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