pi·(7.2/2)^2·x = 2·pi·(7.2/2)^2 + 2·pi·(7.2/2)·x
x = 4.5 = 4 1/2
O = V = 1458/25·pi = 58.32·pi
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.
There are 514 gram rice in each container.
Step-by-step explanation:
Given,
Amount of rice = 6 kg
1 kg = 1000 g
6 kg = 6*1000 = 6000 g
Number of containers = 10
Rice left = 860 g
Let,
x be the amount in each container.
Total amount of rice - Number of containers*Amount in each container = Amount of rice left

Dividing both sides by 10

There are 514 gram rice in each container.
Keywords: linear equation, division
Learn more about division at:
#LearnwithBrainly
Answer:
a) A student travelling to school on public transport: 15/52 or 0.231
b) A student walking to school: 16/52 or 0.308
c) A student not cycling to school: 43/52 or 0.827
Step-by-step explanation:
Total people = 52
Travel Method Frequency
Public Transport 12
Car 15
Cycle 9
Walk 16
Find the relative frequency of.
The formula used will be: 
a) A student travelling to school on public transport:
Given Frequency: 12
Size of sample space: 52
Apply formula: 
Fraction = 12/52
Decimal = 0.231
b) A student walking to school
Given Frequency: 16
Size of sample space. 52
Apply formula: 
Fraction = 16/52
Decimal = 0.308
c) A student not cycling to school.
We will consider all students except those who cycle.
12+15+16 = 43
Given Frequency: 43
Size of sample space. 52
Apply formula: 
Fraction = 43/52
Decimal = 0.827
Answer:
x=−2
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
−3x−32=−2(5−4x)
−3x+−32=(−2)(5)+(−2)(−4x)(Distribute)
−3x+−32=−10+8x
−3x−32=8x−10
Step 2: Subtract 8x from both sides.
−3x−32−8x=8x−10−8x
−11x−32=−10
Step 3: Add 32 to both sides.
−11x−32+32=−10+32
−11x=22
Step 4: Divide both sides by -11.
−11x
−11
=
22
−11