Select all the lines that have a slope of 5/2. Click Image to Draw:
A. A
ΤΑ
B
I
В. В
C. C
D. D
E. E
Explain Your Reasoning:

Lines A and C are having a slope of (5/2) and line E may have the slope of (5/2).

Step-by-step explanation:

Given information:

Line A,B,C,D and E having different slopes;

Now as we know that equation of the line is:

where, m is the slope of the line

Now, Also

Hence, the slope of the line can also be calculated according to that.

Now, as given in the graph:

Line A is having

Hence slope of line A

Similarly:

Line B is having slope of (4/2)

Line C is having slope of (5/2)

Line D is not having slope of (5/2)

And Line E may have the slope of (5/2).

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A rate of 5/2 means that:

If we move 2 units to the right, we move 5 units up.

Let's see all the graphs, the easier way to know this is:

In any point of the line, move to the right 2 units.

Then from that point, move up until you meet again with the line, if you moved 5 units, then the slope is 5/2.

A) The slope is 5/2.

B) The slope is 4/2

C) the slope is 5/2

I can not see very well graphs E and D in the image.

E) may have a slope of 5/2, you should measure it.

D has not a slope of 5/2,  you can compare it with A and see that is totally different, so you can discard option D.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).