The line plot below shows the amount of cereal shyanne ate in 5 days. What is the total number of cups of cereal that shyanne ate in the 5 days

D. 2 1/4 total number of cups of cereals

Step-by-step explanation:

The question lacks diagram. Find the diagram in the attachment.

The total number of cups of cereals that shyanne ate in the 5 days will be individual sum of the fractions of cereals ate during day 2, 3 and 4. X shows the number of times she ate cereal each day.

Day 0 = 0 (she doesn't eat cereal during this day)

Day 1 = 1/4 + 1/4 = 1/2 cup of cereal (X appeared twice). She ate 1/4 cups of cereals twice this day.

Day 2 = 2/4 + 2/4 = 1 cup of cereal (X appeared twice). She ate 2/4 cups of cereals twice this day.

Day 3 = 3/4 cup of cereal (X appeared once). She ate 3/4 cup of cereals once this day.

Day 4 = no cup of cereal was eaten.

Total cups of cereal for the 5days= 0+1/2+1+3/4+0

= 3/2+3/4

= (6+3)/4

= 9/4

= 2 1/4

= 2 1/4 cups of cereal.

Total number of cups of cereal Shyanne ate in 5 days:   or

Step-by-step explanation:

Line plot which shows the amount of cereal Shyanne ate in 5 days in attached below.

We can see that.

Out of 5 days:

2 days, she ate 1/4 cups of cereal

Which means a total of 1/4 + 1/4 cups of cereal

2 days, she ate 1/2 cups of cereal

Which means a total of 1/2 + 1/2 cups of cereal

1 day, she ate 3/4 cups of cereal

Adding the total number of cups to get the

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}

im pretty sure it would be 5 am, hope that , and give brainliest and guys if this is right then press the button and give 5 stars