Assume that trees are subjected to different levels of carbon dioxide atmosphere with 6% of the trees in a minimal growth condition at 340 parts per million (ppm), 13% at 430 ppm (slow growth), 46% at 560 ppm (moderate growth), and 35% at 630 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees

Mean= 554.4

Standard deviation= 82

Step by step Explanation:

The question want us to calculate a) mean b) standard deviation, while we were given the probability mass function of random variables.

The percentage were of the carbon dioxide were given below;

6% of the trees in a minimal growth condition at 340 = 0.06

13% at 430 ppm = 0.13

46% at 560 ppm = 0.46

35% at 630 ppm= 0.35

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8