
Use long division for dividing
Here 'v' term is missing so we use 0vW
Divide
by v-10 using long division
Step 1: Put 8v^2 at the top
step 2: multiply 8v^2 with v-10 and put it at the bottom
step 3: subtract it
We will get remainder 6
so quotient is 
Remainder is 6
Answer:
the total amount of water supplier per hour to the region within a circle of radius R=110 ( that is from distance r, 0<r<110)
![W(R) = 2\pi [1-(R+1)e^{-R}]](https://tex.z-dn.net/?f=W%28R%29%20%3D%202%5Cpi%20%5B1-%28R%2B1%29e%5E%7B-R%7D%5D)
Step-by-step explanation:
if f(r) describes the water supplied at a distance r , the total amount supplied inside a region that goes from 0 until the circle of radius R, is the sum of all f(r) values from 0 until R, that is the integral value over these limits.
The formula deduction can be found in the attached picture
There is an "r" that multiplies e^-r as result of changing from rectangular coordinates to polar ones.(dx*dy --> r*dr*da)
Answer: -8.5
Step-by-step explanation:
I got it right on edg
Answer:
The correct option is;
c. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.
Step-by-step explanation:
Here we have the values
μ = 275 mL
275.4
276.8
273.9
275
275.8
275.9
276.1
Sum = 1928.9
Mean (Average), = 275.5571429
Standard deviation, s = 0.921696159
We put the null hypothesis as H₀: μ₁ = μ₂
Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂
The t-test formula is as follows;

Plugging in the values, we have,
Test statistic = 1.599292
at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912
Our p-value from the the test statistic = 0.1608723≈ 0.1609
Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.