Asoft-drink machine can be regulated so that it discharges an average of μ oz. per cup. if the ounces of fill are normally distributed with a standard deviation of 0.4 oz., what value should μ be set at so that 98% of 6-oz. cups will not overflow?

μ = 5.068 oz

Step-by-step explanation:

Normal distribution formula to use the table attached

Z = (x - μ)/σ

where μ is mean, σ is standard deviation, Z is on x-axis and x is a desired point.

98% of 6-oz. cups will not overflow means that the area below the curve is equal to 0.49; note that the curve is symmetrical respect zero, so, 98% of the cases relied between the interval (μ - some value) and (μ + some value)].

From table attached, area = 0.49 when Z = 2.33. From data, σ = 0.4 oz and x = 6 oz (maximum capacity of the cup). Isolating x from the formula gives

Z = (x - μ)/σ

2.33 = (6 - μ)/0.4

μ = 6 - 2.33*0.4

μ = 5.068

This means that with a mean of 5 oz and a standard deviation of 0.4 oz, the machine will discharge a maximum of 6 oz in the 98% of the cases.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .