For a certain examination, a score of 58 was 2 standard deviations below the mean, and a score of 98 was 3 standard deviations above the mean. what was the mean score for the examination?

74

Step-by-step explanation:

First all, the Z Score formula is Z=(X-ų)/s

Where:

ų is the mean

X is the score

s is the standard deviation

Z is the number of standard deviation above the mean

From the problem we know that X1= 58 and Z1=-2, X2= 98 and Z2=3, but we don’t have the ų and s. To find them we isolate the ų and work with 2 equations replacing the values we know:

ų = X – Zs

a) ų = 58 – (-2)s = 58 + 2s

b) ų = 98 – 3s

Now we divide by 2 the first equation and divide by 3 the second equation:

a) ų/2 = 29 + s

b) ų/3 = 98/3 – s

Now we add the first equation to the second equation eliminating the s:

ų/2 + ų/3 = 29 + 98/3

(3ų + 2ų)/(2x3) = (3x29 + 98)/3

5ų/6 = 185/3

ų = (185×6)/(3×5) = 74

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation:

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90