Better hearing and better sound censory
i think anyways
We can write the function in terms of y rather than h(x)
so that:
y = 3 (5)^x
A. The rate of change is simply calculated as:
r = (y2 – y1) / (x2 – x1) where r stands for rate
Section A:
rA = [3 (5)^1 – 3 (5)^0] / (1 – 0)
rA = 12
Section B:
rB = [3 (5)^3 – 3 (5)^2] / (3 – 2)
rB = 300
B. We take the ratio of rB / rA:
rB/rA = 300 / 12
rB/rA = 25
So we see that the rate of change of section B is 25
times greater than A
Answer:
This is true.
Step-by-step explanation:
Step 1: see that √27 is the same as √3*3*3 and thus 3√3
Step 2: remove parenthesis so that 2(1-3√3) becomes 2 - 6√3
Step 3: observe that the equation now becomes 6√3 + 2 - 6√3 = 2
if you simplify any further, you are left with 0=0, which is true.
Comment if there is any step that requires more details...
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:

And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
For lines to be perpendicular they would need to have opposite slopes, so for a slope of -3/7 the slope of a perpendicular line would be 7/3