Ok, here we go. Pay attention. The formula for the arc length is

. That means that to use that formula we have to find the derivative of the function and square it. Our function is y = 4x-5, so y'=4. Our formula now, filled in accordingly, is

(that 1 is supposed to be negative; not sure if it is til I post the final answer). After the simplification we have the integral from -1 to 2 of

. Integrating that we have

from -1 to 2.

gives us

. Now we need to do the distance formula with this. But we need 2 coordinates for that. Our bounds are x=-1 and x=2. We will fill those x values in to the function and solve for y. When x = -1, y=4(-1)-5 and y = -9. So the point is (-1, -9). Doing the same with x = 2, y=4(2)-5 and y = 3. So the point is (2, 3). Use those in the distance formula accordingly:

which simplifies to

. The square root of 153 can be simplified into the square root of 9*17. Pulling out the perfect square of 9 as a 3 leaves us with

. And there you go!
Answer:
0.35 is the mean of the sampling distribution of the proportion.
Step-by-step explanation:
We are given the following in the question:
Percentage of voters who voted for recall = 35%

Sample size, n = 3150
We have to find the mean of the sampling distribution.
Formula for mean of sampling distribution:

Putting values, we get,

Thus, 0.35 is the mean of the sampling distribution of people sampled in an exit poll who voted for the recall.
Answer: 79 easy do 14x5 then add 9 which equals 79
He makes it up by 1/3=.333%=$75*1.333=$99.98
he then discounts it by 20%=1/5=99.98/5=$19.99
If the original profit is $99.98-$75=$24.98
and then discounted by $19.99
it would be $24.98-$19.99
=
$4.99
Answer:
Mean =
Standard deviation =
Step-by-step explanation:
x P(x)

33 0.02 0.66 21.78
34 0.06 2.04 69.36
35 0.1 3.5 122.5
36 0.2 7.2 259.2
37 0.24 8.88 328.56
38 0.26 9.88 375.44
39 0.1 3.9 152.1
40 0.02 0.8 32
1 36.86 1360.9
We are supposed to find mean and standard deviation
Mean =
Standard deviation =