Ok thank you I think you answered in the wrong place
Answer:
y = 3/2x by making use of angle relationships in triangles
Step-by-step explanation:
Here's one way to solve it.
∠ADE is an external angle to ΔBDE. As such, its measure will be the sum of the measures of the remote interior angles, ∠DBE and ∠DEB:
∠ADE = 2x° +y°
__
If we call the intersection point of AC and DE point G, then ∠AGE is an exterior angle to ΔADG. As such, its measure is the sum of the remote interior angles:
∠AGE = ∠GAD +∠GDA
3y° = x° +(2x° +y°)
2y = 3x . . . . . . . . . . subtract y°, collect terms, divide by °
y = (3/2)x . . . . . . . . divide by 2
Answer:
n=-12
Step-by-step explanation:
We are given that the
operation of all circuits is independent with each other, therefore we can use
the multiplication rule for independent events, which states that P (intersection
of A and B) = P(A) * P(B). In this case, we want the intersection of circuit 1 to
be working with the intersection of circuit 2 on and on until circuit 40. That
is, we want every circuit to work with each other. The given probability that
circuit 1 works is .99. The probability that circuit 2 works is still .99 since
this is independent events. And we see that the probability for each of the 40
circuit to work is .99. <span>
So P (intersection of 1 through 40) = .99 * .99 *
.99.....*.99 = (.99)^40 = .6689717586</span>
Answer:
<span>There is a 0.67 probability
(or 67%) that the product will work.</span>