5 21. The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutiv

Question

5
21. The average of 5 consecutive integers starting with m
as the first integer is n. What is the average of 9
consecutive integers that start with m+2?
A) m + 4
B) n + 6
C) n+ 4
D) m + 5

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Nem 5 months 2021-08-18T14:28:51+00:00 1 Answers 87 views 0

Answers ( )

    0
    2021-08-18T14:30:09+00:00

    Answer:

    n + 4

    Step-by-step explanation:

    Given

    ⅕(m + m + 1 + m + 2 + m + 3 + m + 4) = n

    Required

    ⅑(m + 2 + m + 3 +…….+ m + 10)

    We have:

    ⅕(m + m + 1 + m + 2 + m + 3 + m + 4) = n

    Multiply both sides by 5

    m + m + 1 + m + 2 + m + 3 + m + 4 = 5n

    Collect like terms

    m + m + m + m + m = 5n – 1 – 2 – 3 – 4

    5m = 5n – 10

    Divide both sides by 5

    m = n – 2

    So, we have:

    ⅑(m + 2 + m + 3 + m + 4 + m + 5 + m + 6 + m + 7 + m + 8 + m + 9 + m + 10)

    Collect like terms

    = ⅑(m+m+m+m+m+m+m+m+m+2+3+4+5+6+7+8+9+10)

    = ⅑(9m + 54)

    = m + 6

    Substitute n – 2 for m

    = n – 2 + 6

    = n + 4

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