A scientist planted seeds in 4 sections of soil for an experiment. Not all of the seeda grew into plants. After 20 days, the scientist counted the number of plants in each of the 4 sections. The results are as ahown in the table. Section 1 , sq ft 25, #of plants planted 13. Section 2, SQ ft 100 number of plants planted 38. Section 3, Sq ft 125, # of plants planted 47. Section 4, sq ft 150, # of plants planted 62. Use the data to determine approximately how many plants grew per sq ft. Explain or show how you determined your approximation. Let y be the number of plants expected to grow in x sq ft. Write an equation the scientist could use to model the relationship between y and x.

c

step-by-step explanation:

5 pm

step-by-step explanation:

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.