The scatter plot shows the number of strawberries that have been picked on the farm during the month of February: Part A: Using computer software, a correlation coefficient of r = 0.01 was calculated. Based on the scatter plot, is that an accurate value for this data? Why or why not? (5 points)

Part B: Instead of comparing the number of strawberries picked and the day in February, write a scenario that would be a causal relationship for strawberries picked on the farm. (5 points)

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %