$(sec x - tan x)^2 \\ \\ = sec^2x - 2 sec x tan x + tan^2 x \\ \\ =(1+tan^2 x) - 2 sec x tan x +tan^2 x \\ \\ =1 - 2 sec x tan x + 2 tan^2 x \\ \\ = 1 - 2tan x(sec x - tan x) \\ \\ =1 - \frac{2 sin x}{cos x} (\frac{1-sin x}{cos x}) \\ \\ = 1 - \frac{2 sin x (1-sin x)}{cos^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{1-sin^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{(1-sin x)(1+sin x)} \\ \\ =1-\frac{2 sin x}{1+sin x}$

3 0

3 years ago

A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population

NikAS [45]

Answer:

<u>The probability that the average weight of the boxes will exceed 94 pounds is 43.25%</u>

Step-by-step explanation:

1. Let's review the information provided to answer the question correctly:

Mean of the population of box weights = 90 pounds

Standard deviation of the population of box weights = 24 pounds

Sample size = 36 boxes

2. What is the probability that the average weight of the boxes will exceed 94 lb?

For answering this question, we will use the z-scores table.

For using the correct z-score, we should use the correct standard deviation. In this case, we have that:

94 - 90 = 4 pounds above the mean.

4 pounds are 4/24 of the standard deviation,

therefore, simplifying:

4/24 = 1/6 = 0.1666......and we'll round to 0.17

z-score is 0.17 because 4 pounds is 1/6 of the standard deviation (24).

The probability of a z-score is 0.5675, but in this case we're asked by a weight over 94 pounds, not under 94 pounds, thus:

1 - 0.5675 = 0.4325

<u>The probability that the average weight of the boxes will exceed 94 pounds is 43.25%</u>

8 0

2 years ago