Answer:
(a×b)×c = 0 means that vector c parallel to vector (a×b). But because (a × b) ⊥ a and (a -b) ⊥ b so
c ⊥ a and c ⊥ b.
Hope this helps, have a nice day! :D
We need to work out the z-value of each question to work out the probability
Question a)
We have
X = 5 minutes
The mean, μ = 6.7 minutes
Standard deviation, σ = 2.2 minutes
z-score = (X - μ) / σ = (5 - 6.7) / 2.2 = -0.77
We want to find the probability of z < -0.77 so we read the z-table for the value of z on the left of -0.77 we have the probability P(z< -0.77) = 0.2206
The table is attached in picture 1 below
The probability of assembly time less than 5 minutes is 0.2206 = 22.06%
Question b)
z-score = (10-5) / 2.2 = 2.27
Reading the z-table for P(z<2.27) = 0.9884
The table reading is shown in the second picture below
The probability the assembly time will be less than 10 minutes is 0.9884
We can use this information to find the probability of the assembly time will be more than 10 minutes = 1 - 0.9884 = 0.0116 = 1.16%
Question c)
The value of z between 5 minutes and 10 minutes is
P(-0.77<z<2.27) = P(z<2.27) - P(z< -0.77)
P(-0.77<z<2.27) = 0.9884 - 0.2206 = 0.7678 = 76.78%
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year