Find the value of x

Brooke is 3 years younger than Samantha. In 5 years, the sum of their ages will be 29. Find Brooke's age

Lera25 [3.4K]

Brooke is 8 years old! This is what I did⬇️
5x2=10. (In 5 yrs 10 will add to the sum of their age.
29-10=19. (The sum of their ages right now is 19)
11 + 8 = 19 ( 11 is 3 more than 8 and it equals their ages together)
Brooke:8
Samantha:11

6 0

3 years ago

Read 2 more answers

Barbara is making candy bags with her birthday party she has 24 lollipops, 12 candy bars, and 42 pieces of gum she wants each ba

Arisa [49]

She can make about 5 bags and there will be 4 lollipops, 2 candy bars, and 7 pieces of gum in each bag.

So this is how to solve it:
Basically what you’re doing here is called finding the gcf, or greatest common factor. What you have to do is find what number can divide by 24, 12, and 42; so I chose 2.
After dividing all these numbers by two you get 12, 6, and 21.
3 can divide by all these numbers
After dividing by three you get 4(lollipops), 2(candy bars) and 7(pieces of gum)
Since you can not divide by one number for all of these numbers, 4, 2 and 7 are your final answers.

To find how many’s bags you can make, the answer is 5 since you divided by 2 and 3 and you add them together.

I hope I explained this well! Good luck :D

3 0

3 years ago

Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective

soldier1979 [14.2K]

Answer:

a) $P[C]=p^n$

b) $P[M]=p^{8n}(9-8p^n)$

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:

$P[C]=p^n$

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:

$P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)$

c)

$P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9$

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:

$P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2$

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.