No incorrect. Would only be the origin at 1/2 G if H was (6,7) and I was ( -6,-7)
Using the probability concept, we have that:
a) It would not be unusual to observe one component fail, since the probability that one component fails is greater than 0.05.
b) It would be unusual to observe two components fail, since the probability that two components fail is less than 0.05.
<h3>What is a probability?</h3>
A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>. If a probability is less than 0.05, the event is considered unusual.
In this problem, the probabilities are given as follows:
- 0.21 probability that one component fails, hence not unusual.
- (0.21)² = 0.0441 probability that two components fail, hence unusual.
More can be learned about probabilities at brainly.com/question/14398287
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. the first one of figohohl
Answer: 15
Step-by-step explanation:
(5-1)^2 - (6-5)^3
(4)^2 - (1)^3
16-1
=15
Hi there!
We can calculate dy/dx using implicit differentiation:
xy + y² = 6
Differentiate both sides. Remember to use the Product Rule for the "xy" term:
(1)y + x(dy/dx) + 2y(dy/dx) = 0
Move y to the opposite side:
x(dy/dx) + 2y(dy/dx) = -y
Factor out dy/dx:
dy/dx(x + 2y) = -y
Divide both sides by x + 2y:
dy/dx = -y/x + 2y
We need both x and y to find dy/dx, so plug in the given value of x into the original equation:
-1(y) + y² = 6
-y + y² = 6
y² - y - 6 = 0
(y - 3)(y + 2) = 0
Thus, y = -2 and 3.
We can calculate dy/dx at each point:
At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.
At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.
