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.
Answer:
y 
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 
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)
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.
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