25 POINTS I WILL GIVE BRAINLIEST Patty’s Parasols recorded the monthly rainfall and their umbrella sales for an entire year. The following table shows the inches of rainfall and the number of umbrellas the store sold each month. The relationship between the inches of rainfall and umbrella sales can be modeled using a linear function. Month Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sept. Oct. Nov. Dec. Rainfall (in.) 2.5 3.1 4.0 4.3 3.6 2.4 1.2 0.6 1.5 1.4 1.1 2.6 Umbrella sales 132 140 162 170 156 103 98 87 84 78 110 125 (a) Using technology, determine the correlation coefficient for the data to the nearest hundredth. Explain the steps you used. (b) What does the correlation coefficient for the data indicate about the direction and strength of the linear association between the monthly rainfall and the number of umbrellas sold?

a) From the chart crated with Microsoft Excel, we have that the correlation coefficient, r = √0.8581 ≈ 0.93 to the nearest hundredth

The steps used includes

1) Inputting the given data into the cells on a Microsoft Excel spread sheet

2) Highlighting and sorting the data in the cells in order of increasing Rainfall

3) Generating a dot plot using the sorted data from above

4) Adding the trend line, Square of the linear regression, and the trend line equation

5) Adding the axis labels

(b) The correlation coefficient states that there is a strong positive correlation between the monthly rainfall and and Umbrella sales

Step-by-step explanation:

answer: 457.14

step-by-step explanation:

205 - 45 = 160

160 / .35 = 457.142857

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}