100
ggbnjj mj ddcnmikiyewazxbnoorr
Since there are only two sides (heads & tails) to a coin:
Probability (as fraction) - 1/2
Probability (as percent) - 1/2 x 100 = 50%
Probability (as decimal) - 1 x 50/2 x 50 = 50/100 = 0.5
(P.S. Please mark this answer as the brainliest answer... Thank You)
Answer:
Terms (Variables) = x , d . Their corresponding coefficients = n^2 , 1/2 . Constant = 6
Step-by-step explanation:
6 + n x n + 1/2d
6 + n^2 x + 1/2d
Terms (Variables) = x , d . Their corresponding coefficients = n^2 , 1/2 . Constant = 6
Answer: The mean and standard deviation are 567.2 and 89.88 resp.
Step-by-step explanation:
Since we have given that
For 370 parts per million = 7% = 0.07
For 440 parts per million = 10% = 0.10
For 550 parts per million = 49% = 0.49
For 670 parts per million = 34% = 0.34
So, Mean of the carbon dioxide atmosphere for these trees would be
![E[x]=370\times 0.07+440\times 0.1+550\times 0.49+670\times 0.34=567.2](https://tex.z-dn.net/?f=E%5Bx%5D%3D370%5Ctimes%200.07%2B440%5Ctimes%200.1%2B550%5Ctimes%200.49%2B670%5Ctimes%200.34%3D567.2)
And
![E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 0.34=329794](https://tex.z-dn.net/?f=E%5Bx%5E2%5D%3D370%5E2%5Ctimes%200.07%2B440%5E2%5Ctimes%200.1%2B550%5E2%5Ctimes%200.49%2B670%5E2%5Ctimes%200.34%3D329794)
So, Variance would be
![Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16](https://tex.z-dn.net/?f=Var%5C%20x%3DE%5Bx%5E2%5D-E%5Bx%5D%5E2%3D329794-567.2%5E2%3D8078.16)
So, the standard deviation would be

Hence, the mean and standard deviation are 567.2 and 89.88 resp.
Step-by-step explanation:
1 t=16/21
2.m=2
3.n=13/7
4.a=2
5.x=6/17
6.x=15
7.s=21/4
8. t=7/3
9. s=1
10. s=6/61
11. x=1/3
12. r=27/16
13. c=−1
14.m=9/5n
15. j=−117/58