Acandy company produces individually wrapped candies. the quality control manager for the company believes that the weight of the candies is approximately normally distributed with mean 720 milligrams (mg). if the manager’s belief is correct, which of the following intervals of weights will contain the largest proportion of the candies in the distribution of weights?

a) 740 mg to 780 mg
b) 700 mg to 740 mg
c) 680 mg to 720 mg
d) 660 mg to 720 mg
e) 620 mg to 660 mg

Option B.

Step-by-step explanation:

It is given that the weight of candies is approximately normally distributed with mean 720 milligrams (mg).

It means most of the candies weight lies around the 720 milligrams.

In normal distribution,

Interval = Mean ± n × (standard deviation)

Where, n=1 for 68%, n=2 for 95% and n=3 for 99.7% and mean is the midpoint of the interval.

From the given option, only option B is true because 720 mg is the midpoint of 700 mg to 740 mg.

Therefore, the correct option is B.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

i think it is 70

step-by-step explanation: