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LUCKY_DIMON
1 month ago
10

The failure rate for a guided missile control system is 1 in 1,000. Suppose that a duplicate but completely independent control

system is installed in each missile so that, if the first fails, the second can take over. The reliability of a missile is the probability that it does not fail. What is the reliability of the modified missile
Mathematics
1 answer:
Inessa [12.5K]1 month ago
6 0

Answer:

The reliability is approximately 0.999999.

Step-by-step explanation:

The control system for a guided missile has a failure rate of 1 in every 1,000.

Therefore, the likelihood of the missile functioning successfully is \frac{1000 - 1}{1000} = \frac{999}{1000}.

In the newly structured system, if the first one fails, the backup can take over.

There are two outcomes possible: either the first fails or it works.

If the first is successful, the probability is \frac{999}{1000} = 0.999.

If the first one fails, the probability becomes \frac{999}{1000} \times\frac{1}{1000} = 0.000999.

The overall reliability of the modified missile is 0.999 + 0.000999 = 0.999999.

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Sketch the region bounded by the curves y=3x3, x+y=4, and y=0. Find the coordinates of the centroid.
AnnZ [12381]
The attached graph illustrates the region. The centroid's coordinates are (5/3, 1). The centroid's coordinates are determined by averaging the coordinates of the area; Oₓ = (Aₓ+Bₓ+Cₓ)/3 = (0+1+4)/3 = 5/3 and O(y) = (A(y) + B(y) + C(y)) = (0+3+0)/3=3/3=1.
5 0
1 month ago
Which is the solution of the quadratic equation (4y – 3)2 = 72? y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction a
Inessa [12570]

Answer:

y = \frac{3 + 6\sqrt{2} }{4} y y = \frac{3 - 6\sqrt{2} }{4}

y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction

Explicación paso a paso:

La ecuación cuadrática que tenemos es (4y - 3)² = 72

Debemos encontrar el valor de y.

Ahora, 4y - 3 = ± 6√2

⇒ 4y = 3 ± 6√2

⇒ y = \frac{3 + 6\sqrt{2} }{4} y y = \frac{3 - 6\sqrt{2} }{4}

Por lo tanto, las soluciones son y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction (Respuesta)

6 0
1 month ago
Read 2 more answers
: Xem xét 2 nhà đầu tư A và B với đường cầu về cổ phiếu như sau: Nhà đầu tư A: 2P = 300 – 2Q Nhà đầu tư B: P = 150 – 3Q a. Tại m
Zina [12379]

Đáp án:

Giải thích từng bước:

5 0
1 month ago
Heights of adult males are known to have a normal distribution. a researcher claims to have randomly selected adult males and me
Leona [12618]

Answer:

The total probability exceeds 100%, indicating a problem with the findings; moreover, the distribution shows excessive uniformity which disqualifies it as a normal distribution.

Detailed explanation:

The sum of probabilities should be exactly 100%. When you add the probabilities of this distribution:

22+24+21+26+28 = 46+21+26+28 = 67+26+28 = 93+28 = 121

This exceeds 100%, highlighting a significant error in the results.

A typical normal distribution possesses a bell curve. If we plot the probabilities for this distribution, we'd see bars at 22, 24, 21, 26, and 28.

The bars would fail to form a bell-shaped curve, confirming that this is not a normal distribution.

3 0
1 month ago
Read 2 more answers
If 6 components are drawn at random from the container, the probability that at least 4 are not defective is . If 8 components a
Svet_ta [12734]
The question is missing some information. It should be phrased as follows:

<span><span>A container has 50 electronic components, with 10 identified as defective. If 6 components are randomly selected from the container, what is the probability that at least 4 of them are not defective? Additionally, if 8 components are drawn at random from the container, what is the probability that exactly 3 are defective?

</span>Answers
<span>Part 1.  0.02
Part 2. </span></span>0.0375<span><span>

</span>Explanation
Probability denotes the likelihood of an event occurring. It is computed as:
probability = (Number of favorable outcomes)/(Number of total outcomes)

Part 1
When 6 components are chosen, if 4 are confirmed functioning, then 2 must be defective.
P(at least 4 functional) = 4/40</span>× 2/10
                                            = 1/10 × 1/5
                                            = 1/50
                                            = 0.02

Part 2
Choosing 8 components, if 3 are defective, then 5 are functioning.
P(3 defective) = 3/40 × 5/10
                             = 15/400
                             = 3/80
                            = 0.0375
4 0
29 days ago
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