What is the longest diagonal of a rectangular 44 ft times ×33 ft times ×55 ft box?

  55√2 ft

Step-by-step explanation:

The space diagonal is the root of the sum of the squares of the edge lengths:

  √((44 ft)² +(33 ft)² +(55 ft)²) = √(1936 +1089 +3025) ft = √6050 ft

  = 55√2 ft

If you intend 3, 4, and 5, divide the result by 11: 5√2 ft.

The face diagonal of the 3×4 face will have length 5 (remember the 3-4-5 right triangle?). Then that face diagonal and the edge of length 5 are the legs of a right triangle whose hypotenuse is the space diagonal of interest. Since it is an isosceles right triangle, the hypotenuse length is √2 times the edge length, hence 5√2. (If you intend 44×33×55, multiply these numbers by 11.)

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;