<span>(2x + y2)<span>5 heres you answer your so welcome</span></span>
Answer:
8.49 × 10⁻³
Explanation:
Standard form of a number: A. × 10ᵇ
(where 'A' is the number, b is the exponent over 10)
For the number "0.00849", move the decimal point 3 places right so the exponent of 10 is -3.
Standard form: 8.49 × 10⁻³
Other examples of standard/scientific forms are: standard forms
i) 0.043, move the decimal point 2 places right = 4.3 × 10⁻²
ii) 4030, move the decimal point 3 places left = 4.03 × 10³
Answer:
x = 1.71
Step-by-step explanation:
![x^3-20=-15\\x^3=-15+20\\x^3=5\\\sqrt[3]{x^3} =\sqrt[3]{5} \\x=1.71](https://tex.z-dn.net/?f=x%5E3-20%3D-15%5C%5Cx%5E3%3D-15%2B20%5C%5Cx%5E3%3D5%5C%5C%5Csqrt%5B3%5D%7Bx%5E3%7D%20%3D%5Csqrt%5B3%5D%7B5%7D%20%5C%5Cx%3D1.71)
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:
![P[C]=p^n](https://tex.z-dn.net/?f=P%5BC%5D%3Dp%5En)
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.
Answer:
Option D is correct
Step-by-step explanation:
From the question, we have:



Required
The tree diagram
The selection process is as follows:

-- i.e. we write sack 2 as a subset of 1
-- i.e. we write sack 3 as a subset of 2
<em>The only tree diagram that illustrates this is: option D</em>