A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 500 sq ft) and the quality of the resulting lawn. Eight similar plots of land are selected and each is planted with a particular density of seed. One month later the quality of each lawn is rated on a scale of 0 to 100. The sample data are given below. Seed Density Lawn Quality
1 30
1 40
2 40
3 40
3 50
3 65
4 50
5 50

Required:
At the 1% level of significance, is there evidence of an association between seed density and lawn quality?

There is no significant evidence to support the claim that an association between seed density and lawn quality exists.

Step-by-step explanation:

Given the data :

Seed Density Lawn Quality

1 30

1 40

2 40

3 40

3 50

3 65

4 50

5 50

Using technology to obtain the correlation Coefficient between seed density and lawn quality, the Coefficient, R = 0.6

H0 : ρ = 0

H1 : ρ ≠ 0

We can then calculate the test statistic :

Test statistic = r / √(1 - r²) / (n - 2)

Test statistic = 0.6 / √(1 - 0.6²) / (8 - 2)

Test statistic = 0.6 / 0.3265986

Test statistic = 1.837

Pvalue from test statistic : = 0.11583

α = 0.01

Pvalue > α ;

We fail to reject the null

b is 38 degrees.

a is 52 degrees.

this is because if b is 38 and a is 14 degrees more than b than 38+14=52 and 52+38=90 degrees. hope this .

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x