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grandymaker [24]
3 years ago
14

The class midpoint of a class is

Mathematics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

The center of that class, which is the sum of its largest and smallest values divided by 2 ⇒ E

Step-by-step explanation:

* Lets explain what is the class mid-point

- It is defined as the average of the upper and lower class limits

- The class midpoint is the lower class limit plus the upper class limit

 divided by 2

- The easiest way to find the class mid-point is to add the upper

  and lower boundary and divide your answer by two

- The lower limit for every class is the smallest value in that class.

- The upper limit for every class is the greatest value in that class

* <u><em>Lets solve the problem</em></u>

- It is not the largest value of that class minus the class width

- Its not the difference between the largest and smallest values of

 that class

- It is not the difference between the largest and smallest values of

that class divided by 2

- It is the center of that class, which is the sum of its largest and

 smallest values divided by 2

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If the roots of x²-7x+k=0 are m and m-1, find the value of the constant k​
creativ13 [48]

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x = [-(-7) ± √((-7)2 - 4(1)(k))]/2(1)

x = [7 ± √(49 - 4k)]/2

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since the roots are m and m-1,

m = [7 + √(49 - 4k)] / 2 ------------ eqn(1), and

m - 1 = [7 - √(49 - 4k)] / 2 ----------- eqn(2)

add 1 to both sides of eqn(2)

m = [7 - √(49 - 4k)]/2 + 1 --------------- eqn(3)

eqn(1) = eqn(3)

[7 + √(49 - 4k)]/2 = [7 - √(49 - 4k)]/2 + 1

7/2 + (√(49 - 4k))/2 = 7/2 - (√(49 - 4k))/2 + 1

collect like terms,

(√(49 - 4k))/2 + (√(49 - 4k))/2 = 7/2 - 7/2 + 1

(√(49 - 4k))/2 + (√(49 - 4k))/2 = 0 + 1

(√(49 - 4k))/2 + (√(49 - 4k))/2 = 1

multiply through by 2

√(49 - 4k) + √(49 - 4k) = 2

2√(49 - 4k) = 2

divide both side by 2

√(49 - 4k) = 1

square both sides

49 - 4k = 1

collect like terms,

4k = 49 - 1

4k = 48

divide both sides by 4

k = 48/4

k = 12

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