PLEASE HELP ASAP I always give brainliests! You have found the following ages (in years) of 4 bears. Those bears were randomly selected from the 22 bears at your local zoo: 5,4,6,39 Based on your sample, what is the average age of the bears? What is the standard deviation? Round your answers to the nearest tenth.
Average age: years old
Standard deviation: years

first to find average age add all the ages divide by number of ages 4

5+4=9

9+6=15

15+39=54

54/4=13.5

For standard deviation find the mean/average

13.5

then subtract the mean from each number/age

5-13.5=-8.5

4-13.5=-9.5

6-13.5=-7.5

39-13.5=25.5

then square each number

-8.5^2=-72.25

-9.5^2=-90.25

-7.5^2=-56.25

25.5^2=650.25

then find the mean/average of the new numbers

-72.25+-90.25+-56.25+650.25=431.5

431.5/4=107.85

then take the square root of the new mean/average

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10