I need your help on this

AB, CD, and EF intersect at O. If CD is perpendicular to EF and m

REY [17]

Answer:

m(∠AOF) = 148°

Step-by-step explanation:

From the figure attached,

CD intersects line EF at a point O.

Line CD is perpendicular to the line EF.

m(∠AOE) = 32°

m(∠COE) = 90°

Since m(∠COE) = m(∠AOE) + m(∠AOC) = 90°

32° + m(∠AOC) = 90°

m(∠AOC) = 90° - 32° = 58°

m(∠AOF) = m(∠AOC) + m(∠COF)

= 58° + 90°

= 148°

Therefore, m(∠AOF) = 148° will be the answer.

4 0

3 years ago

4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

galben [10]

Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$