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Mathematics

1 answer:

Viktor [21]3 years ago

6 0

Answer:

mean (average) is 17.6

mean absolute deviation is 15.2

Step-by-step explanation:

to find mean, u add up all your numbers and divide by how many there are...

to find mean absolute deviation you find the mean of the data and subtract that amount to each of your data pieces...

then u find absolute value of each of them (to make them all positive)

then u add all of those numbers up and divide by how many there are, and that is your answer!!

Hope this helps u in the future!!

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What is the domain and range of the following functions? Number 1 please

MrMuchimi

Answers:
Part 1 (the ovals)
Domain = {-6,-1,1,5,7}
Range = {-4,-1,2,4}
-------------------
Part 2 (the table)
Domain = {1,-3,-2}
Range = {-2,5,1}
-------------------
Part 3 (the graph)
Domain = {1, 2, 3, 4, 5, 6}
Range = {-1, 0, 1, 2, 3, 6}

===============================================

Explanation:

Part 1 (the ovals)
The domain is the set of input values of a function. The input oval is the one on the left.
All we do is list the numbers in the input oval to get this list: {-6,-1,1,5,7}
The curly braces tell the reader that we're talking about a set of values.
So this is the domain.

The range is the same way but with the output oval on the right side
List those values in the right oval and we have {-4,-1,2,4}
Which is the range. That's all there is to it.

------------------------------

Part 2 (The tables)

Like with the ovals in part 1, we simply list the input values. The x values are the input values. Notice how this list is on the left side to indicate inputs.
So that's why the domain is {1, -3, -2}. Optionally you can sort from smallest to largest if you want. Doing so leads to {-3, -2, 1}

The range is {-2,5,1} for similar reasons. Simply look at the y column

Side Note: we haven't had to do it so far, but if we get duplicate values then we must toss them.
------------------------------
Part 3 (the graph)

Using a pencil, draw vertical lines that lead from each point to the x axis. You'll notice that you touch the x axis at the following numbers: 1, 2, 3, 4, 5, 6
So the domain is the list of those x values (similar to part 2) and it is {1, 2, 3, 4, 5, 6}

Erase your pencil marks from earlier. Draw horizontal lines from each point to the y axis. The horizontal lines will arrive at these y values: -1, 0, 1, 2, 3, 6
So that's why the range is {-1, 0, 1, 2, 3, 6}

5 0

4 years ago

I'm not understanding this can you help me

hoa [83]

Answer:

gawin mo pag isipan mo wag puro brainly

3 0

3 years ago

Read 2 more answers

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:

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

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

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.