Julie wants to compare the heights of two types of tomato plants, Type A and Type B, for her vegetable garden. She thinks the mean height of Type A plants will be greater than the Type B, but she doesn’t know how much greater. She also wants to know about the variability of the heights of each plant type. She uses the data she collects to generate the list below. Part A. Create two dot plots using the same scale to help Julie compare her data sets.

Part B. Julie observes that there is some overlap in the sets. Calculate the mean height of each of these tomato plants. Round your answers to the nearest whole number. What is the difference in the mean height of these tomato plants?

Part C. Compare the variability of the sets. Look at the mean heights and the variability of the sets to make a comparison statement about the two sets.

Use words, numbers, and/or pictures to show your work.

binomial distribution

median= 2

range = 7-0 = 7

how to find median?

no. of term=28

median=(28+1)/2=14.5th term

the 14th and 15th term are 2, therefore the median is (2+2)/2 = 2

1.4

2.3

step-by-step explanation:

the formula for the constant of variation is

y= kx

divide each side by x

y/x =kx/x

y/x =k

we know a and y

1.   y/x = k

4/1 =4

check the other values to make sure they are the same

8/2 = 4   12/3 =4   and 16/4 =4  

the constant of variation is 4

2.   y/x = k

3/1 =3

check the other values to make sure they are the same

6/2 = 3   9/3 =3   and 12/4 =3  

the constant of variation is 3

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}