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.
Your answer will be t = -3P/2Pr + A/4Pr. :)
Answer:
35% of $64.99=35/100 of 64.990=0.35×64.99=$22.7465
DISCOUNTED PRICE=$64.99-$22.7465=$42.2435
PLEASE GIVE BRAINLIEST
Answer:
One-sample Z-test
Step-by-step explanation:
As the sample size is greater than 30, assuming the data is a simple random sample, and that the data has a normal distribution, the sociologist should conduct a one-sample z-test to calculate the p-value.