Answer: The total amount the water change was -28 gallons by the end of the week <u>OR</u> The total amount the water changed decreased 28 gallons by the end of the week. Depending on how you want to word it.
Hope this helps!
Step-by-step explanation:
<u>Create you equation:</u>
x = the total amount the water changed
d = the amount of days
x = -4d
<u />
<u>Insert the known information into the equation and solve:</u>
x = -4d
x = -4(7)
x = -28
$4 is the original price of the jar consisting of peanut butter so therefor it = 100%. It's for sale at $3.60, so we divide $4 by 100% so we get the percentage for each cent and then we times it by 3.60 to get the percentage for $3.60 which is 90%. Now we subtract 90% from 100% to get the discounted percentage for 40cents :)
Answer:
Step-by-step explanation:
If the 4th lies opposite the 11th, then you know the 3rd lies opposite the 10th, the 2nd opposite the 9th, and the 1st opposite the 8th; meanwhile, in the reverse direction you'd find that the 5th lies opposite the 12th, the 6th opposite the 13th, and the 7th opposite the 14th. Move up one more bead and you're back at the 1st, which you already know lies opposite the 8th. Therefore there are 14 total bead
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.