Describe the DOMAIN with Words Describe the domain of the relationship shown here.
1 answer:
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Answer:
- as written, c ≈ 0.000979 or c = 4
- alternate interpretation: c = 0
Step-by-step explanation:
<em>As written</em>, you have an equation that cannot be solved algebraically.
(32^2)c = 8^c
1024c = 8^c
1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero
A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.
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If you mean ...
32^(2c) = 8^c
(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2
2^(10c) = 2^(3c) . . . . . . . simplify
10c = 3c . . . . . . . . . . . . . .log base 2
7c = 0 . . . . . . . . . . . . . . . subtract 3c
c = 0 . . . . . . . . . . . . . . . . divide by 7
Answer:
The probability is 0.971032
Step-by-step explanation:
The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.
The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:
In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:
(eq. 1)
So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:
P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)
We can also calculated that as:
P(x ≥ 5) = 1 - P(x ≤ 4)
Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)
Then, if we calculate every probability using eq. 1, we get:
P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765
P(x ≤ 4) = 0.028968
Finally, P(x ≥ 5) is:
P(x ≥ 5) = 1 - 0.028968
P(x ≥ 5) = 0.971032