In the given diagram, the measure of ∠3 will be 105°.
In the given diagram, ∠3 and ∠6 are consecutive interior angles.
<h3>How to form supplementary angles by transversal?</h3>
If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary.
That is,
∠3 + ∠6 = 180°
From the given information,
∠6 = 75°
Then,
∠3 + 75° = 180°
∠3 = 180° - 75°
∠3 = 105°
Hence, the measure of ∠3 will be 105°.
Learn more about the measures of angles here: brainly.com/question/2883630
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Answer:
5.5 gallons
Step-by-step explanation:
There are 4 quarts in 1 gallon.
22 ÷ 4 = 5.5
I hope this helps :)
1. Use Difference of squares: a^2-b^2=(a+b)(a-b)
(x+k)(x-k)=0
2. Solve for x and k
x=+-k
k=+-x
That would simplify to 32x^5+400x^4+200x^3+5000x^2+6250x+3125
You would do that because you have to write it repeatedly as (2x+5)(2x+5) and so on.
Answer:
a) ![P[C]=p^n](https://tex.z-dn.net/?f=P%5BC%5D%3Dp%5En)
b) ![P[M]=p^{8n}(9-8p^n)](https://tex.z-dn.net/?f=P%5BM%5D%3Dp%5E%7B8n%7D%289-8p%5En%29)
c) n=62
d) n=138
Step-by-step explanation:
Note: "Each chip contains n transistors"
a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:
b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.
We can calculate this as a binomial distribution problem, with n=9 and k≥8:
![P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)](https://tex.z-dn.net/?f=P%5BM%5D%3DP%5BC_9%5D%2BP%5BC_8%5D%5C%5C%5C%5CP%5BM%5D%3D%5Cbinom%7B9%7D%7B9%7DP%5BC%5D%5E9%281-P%5BC%5D%29%5E0%2B%5Cbinom%7B9%7D%7B8%7DP%5BC%5D%5E8%281-P%5BC%5D%29%5E1%5C%5C%5C%5CP%5BM%5D%3DP%5BC%5D%5E9%2B9P%5BC%5D%5E8%281-P%5BC%5D%29%5C%5C%5C%5CP%5BM%5D%3Dp%5E%7B9n%7D%2B9p%5E%7B8n%7D%281-p%5En%29%5C%5C%5C%5CP%5BM%5D%3Dp%5E%7B8n%7D%28p%5E%7Bn%7D%2B9%281-p%5En%29%29%5C%5C%5C%5CP%5BM%5D%3Dp%5E%7B8n%7D%289-8p%5En%29)
c)
![P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9](https://tex.z-dn.net/?f=P%5BM%5D%3D%280.999%29%5E%7B8n%7D%289-8%280.999%29%5En%29%3D0.9)
This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.
d) If the memoty module tolerates 2 defective chips:
![P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2](https://tex.z-dn.net/?f=P%5BM%5D%3DP%5BC_9%5D%2BP%5BC_8%5D%2BP%5BC_7%5D%5C%5C%5C%5CP%5BM%5D%3D%5Cbinom%7B9%7D%7B9%7DP%5BC%5D%5E9%281-P%5BC%5D%29%5E0%2B%5Cbinom%7B9%7D%7B8%7DP%5BC%5D%5E8%281-P%5BC%5D%29%5E1%2B%5Cbinom%7B9%7D%7B7%7DP%5BC%5D%5E7%281-P%5BC%5D%29%5E2%5C%5C%5C%5CP%5BM%5D%3DP%5BC%5D%5E9%2B9P%5BC%5D%5E8%281-P%5BC%5D%29%2B36P%5BC%5D%5E7%281-P%5BC%5D%29%5E2%5C%5C%5C%5CP%5BM%5D%3Dp%5E%7B9n%7D%2B9p%5E%7B8n%7D%281-p%5En%29%2B36p%5E%7B7n%7D%281-p%5En%29%5E2)
We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.
