If the average, mode, and range of 5 different numbers are all 8, find the sum of all possible values of the second largest number.

I believe you meant:

If the average, mode, and range of 5 different numbers are all 8, find the sum of all possible values and the second largest number

Sum of all possible values = 40

Second largest number= 8

Explanation:

Since our mean, mode, and range is 8, our sum is 40 and our second largest number is 8.

The sum of all the numbers is 40 since our mean is 8 and there are 5 numbers(8×5)

To find possible values, we already know the mode is 8 and so we know 8 must appear more than once. If we have 8 twice then we are left with 24(sum is 40). We have 3 numbers left and the range is 8 so we must have a maximum number and minimum number that has a difference of 8 and a third number, all adding up to 24(sum left). Within 24, only 12-4+8 satisfies the condition, hence our numbers are: 4,8,8,8,12.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

heyyy

step-by-step explanation: