Answer:
3 1/3 is your answer.
Step-by-step explanation:
Rewriting equation:
4 + 5/6 - 1 - 1/2
4 - 1 = 3
5/6 - 1/2 = ?
Find the LCD of 5/6 and 1/2 and rewrite to solve.
5/6 - 3/6 = 2/6
2/6 = 1/3
3 + 1/3 = 3 1/3
Whole number: 3
numerator: 1
denominator: 3
Based on the interest rate and continuous compounding, the investment would double in value after 18.5 years.
We have given that,
investment to double at a 3 3/4% interest rate,
<h3>When will the investment double in value?</h3>
The future value using continuous compounding is:
= Amount x e ^ (rate x time)
Interest is
= 3.75%
<h3>What is the formula of an exponential function?</h3>
2 = e ^ (0.0375 x time)
In2 = 0.0375 x time
t = In2 / 0.0375
t= 18.5 years
To learn more about the compounded continuously visit:
brainly.com/question/16731646.
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Since d x pi = circumference
I use 3.14 as pie
I did 9 x pi = 28.26 I DID NOT divide this by two because there are three half circles and 28.26 counts as two of them
Since there is another half circle I divide 28.26 by 2
28.26 + 14.13 = 42.39
Now plus the bottom length
42.39 + 9 = 51.39
Brainliest answer please?
Answer:
The answer is D: the number of candies in a box.
Step-by-step explanation:
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.
