A set of 9 numbers has a mean of 20. What additional number must be included in this set to create a new set with a mean that is 4 less than the mean of the original set?

Greetings from Brasil...

The average for a set of 9 elements will be

(A + B + C + D + E + F + G + H + I) ÷ 9 = 20

Let's make (A + B + C + D + E + F + G + H + I) like S

(I chose S to remember a sum)

Let us think

S ÷ 9 = 20

S = 20 × 9

S = 180

So,  (A + B + C + D + E + F + G + H + I) = 180

According to the statement, we will include a number (element J) in the sum to obtain a mean of (20 - 4), that is:

as seen above, (A + B + C + D + E + F + G + H + I) = 180, then

(180 + J) ÷ 10 = 16

(180 + J) = 160

J = 160 - 180

So, including the number - 20 (minus 20) in the original mean we will obtain a new mean whose result will be 16

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.