Part 1 of 3 The weights of a certain brand of candies are normally distributed with a mean weight of 0.8591 g and a standard deviation of 0.0513 g. A sample of these candies came from a package containing 461 candies, and the package label stated that the net weight is 393.5 g. (If every package has 461 candies, the mean weight of the candies must exceed
393.5
461=0.8535 g for the net contents to weigh at least 393.5 g.)
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8535 g.
The probability is
(Round to four decimal places as needed.)

we need to calculate a z-score and refer to the normal distribution tables.

Remember that not all distribution tables read the same way, but they should show you by a diagram whether the percentage given is for the area to the left of the z or to the right.

Z = (X - μ)/σ      or       Z = (.8543 - .8605)/.0513 = -.1209

The area to the right of this Z-score translates to a probability of 0.5478

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°