The probability that your homework is on disc i is 12.25%.
<h3><u>Probabilities </u></h3>
Given that the disc containing the only copy of your homework just got corrupted, and the disk got mixed up with two other corrupted discs that were lying around, and it is equally likely that any of the three discs holds the corrupted remains of your homework, and your computer expert friend offers to have a look at one of the discs, and you know from past experience that his probability of finding your homework from a disc is 0. 35, given that the homework is in there, given that he searches on disc 1 but cannot find your homework, to determine what is the probability that your homework is on disc i, for i = 1, 2, 3, the following calculation must be performed:
- Disc i = 0.35 x 0.35
- Disc i = 0.35^2
- Disc i = 0.1225
- 0.1225 x 100 = 12.25
Therefore, the probability that your homework is on disc i is 12.25%.
Learn more about probability in brainly.com/question/26444510
Answer:
What Are the Answer Choices
Explanation:
I believe I know the answer, but I need the answer choices.
Answer: do you still need help with it?
Explanation:
The net electric field is the vector sum of the components of the electric
field produced by the two charges.
The values of the magnitude and direction of the net electric field at the origin (approximate values) are;
- 12.6 ° above the negative x–axis
<h3>How are the net electric field magnitude and direction calculated?</h3>
The possible questions based on a similar question posted online are;
(a) The net electric field at the origin.
The electric field due to charge q₁ is given as follows;
Which gives;
Which gives;
Therefore;
![\vec {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)](https://tex.z-dn.net/?f=%5Cvec%20%20%7BE%7D%20%3D%20%5Cleft%5B%2021.6%20%5C%2C%20N%2FC%20-%20150%20%5C%2C%20N%2FC%20%5Cright%5D%20%5Cleft%28%20%5Chat%20x%20%5Cright%29%20%2B%20%5Cleft%2828.8%20%5C%2C%20N%2FC%20%5Cright%29%20%5Cleft%28%20%5Chat%20y%20%5Cright%29)
The magnitude of the net electric field is therefore;
E = 
≈ 131.6
- The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>
(b) The direction of the net electric field at the origin.
Learn more about electric field strength here:
brainly.com/question/14743939
brainly.com/question/3591946
