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den301095 [7]
3 years ago
13

I need help plss help

Mathematics
1 answer:
kirill [66]3 years ago
6 0
Need help with what ...?.???????
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Angelina factored (x - 48 ) and wrote that it was equal to (x^2 - 4^2)^2(x^2 + 4^2). Use complete sentences to explain how you c
defon
(x^2 - 4^2)^2 = x^4 - 4^4
(x^4 - 4^4) (x^2 + 4^2) = x^2 - 4^6
The Angela's solution is wrong cause when you multiply you get a different answer.
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3 years ago
Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B123%7D%7B65%7D" id="TexFormula1" title="\frac{123}{65}" alt="\frac{123}{65}"
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Answer: 1.89230769231

Step-by-step explanation:

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3 years ago
I’ve tried to work this out but I can’t get it figured out. Can someone help me
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G(2) = -4(2) = -8
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3 years ago
13. Tom has $14 more than you do.
puteri [66]

Answer:

46

Step-by-step explanation:

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8 0
3 years ago
A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has four identi
Tcecarenko [31]

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = \dbinom{4}{0} × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = \dbinom{4}{1} × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

4 0
2 years ago
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