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Hunter-Best [27]
3 years ago
15

Sketxh the graph of y=(x+3)^3 and state the points where the graph crosses the axes​

Mathematics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer: the graph crosses the x-axis at x = -3

<u>Step-by-step explanation:</u>

y = (x + 3)³

To find where the graph crosses the x-axis, let y = 0 and solve for x:

0 = (x + 3)³

0 = (x + 3)    with a multiplicity of 3

-3 = x           with a multiplicity of 3.

Since multiplicity is an ODD number, the graph CROSSES the x-axis at x = -3

<em />

<u>Graph:</u>

  • Leading coefficient is POSITIVE so right side goes to +∞
  • Degree of polynomial is ODD so left side goes to -∞

<em>graph is attached</em>

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A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has eight ident
Fudgin [204]

Answer:

a) 0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b) 0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either they last more than 1,000 hours, or they do not. Components operate independently, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

One subsystem has eight identical components, each with a probability of 0.1 of failing in less than 1,000 hours.

So 1 - 0.1 = 0.9 probability of working for more, which means that p = 0.9

a. exactly two of the four components last longer than 1000 hours.

This is P(X = 2) when n = 4. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.9)^{2}.(0.1)^{2} = 0.0486

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b. the subsystem operates longer than 1000 hours.

The subsystem has 8 components, which means that n = 8

It will operate if at least 4 components are working correctly, so we want:

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{8,0}.(0.9)^{0}.(0.1)^{8} \approx 0

P(X = 1) = C_{8,1}.(0.9)^{1}.(0.1)^{7} \approx 0tex][tex]P(X = 2) = C_{8,2}.(0.9)^{2}.(0.1)^{6} \approx 0

P(X = 3) = C_{8,3}.(0.9)^{3}.(0.1)^{5} = 0.0004

Then

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0 + 0 + 0.0004 = 0.0004

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0004 = 0.9996

0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

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Answer:

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Answer:

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