Answer:
I = $1,200.00
Step-by-step explanation:
First, converting R percent to r a decimal
r = R/100 = 6%/100 = 0.06 per year,
then, solving our equation
I = 4000 × 0.06 × 5 = 1200
I = $1,200.00
Answer:
(6, 2), (0, -2), (-3, -4)
Step-by-step explanation:
The intercepts are clearly (0, -2) and (3, 0). This eliminates the bottom two choices.
Both of these are listed in the upper left choice, but the third point there, (-6, -5), is not on the graph. This leaves only the upper right choice. Checking the remaining two points in that choice, (6, 2) and (-3, -4), confirms they are both on the graph.
What is the question pls so I might be able to help
Answer:
5.9987174899
Step-by-step explanation:
9.654km / 1.609344
= 5.9987174899mi
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.
