In a cannery, assembly lines I, II, and III account for 40, 40, and 20 percent of the total output. If 0.4 percent of the cans from assembly line I are improperly sealed, and the corresponding percentages from assembly line II and III are 0.3 percent and 0.3 percent, what is the probability P that a can produced by this cannery will be improperly sealed, and what is the probability Q that an improperly sealed can will have come from assembly line I?a. P = 0.0034; Q = 0.7353.

b. P = 0.00755; Q = 0.3434.

c. P = 0.006875; Q = 0.3434.

d. P = 0.00825; Q = 0.364.

e. P = 0.006875; Q = 0.3636.

the solution is 12x-6y

x=half

step-by-step explanation:

c = 0.429t

step-by-step explanation:

the constant rate at which carlos harvests cassava is 35 minutes per 15 cassavas 15/35 = 0.429 cassava per minute

therefore, an equation that models the number of cassava (c) that harvest carlos in a time (t) is the following:

where t is the time in minutes.

this means that when 35 minutes have elapsed 15 cassava have been harvested, at 70 minutes 30 cassava have been harvested and so on.

2/4 is the answer

step-by-step explanation:

x= total pieces

both ate 9 pieces

mindy ate 3 so troy ate 6

6= (1/4)x

x =24