<span>Denote x as the interval, then:
186 = 50 + 3 + (3+x) + (3+2x) + (3+3x) + (3+4x) + (3+5x) + (3+6x) + (3+7x)
186 = 74 + 28x
x = 4
Age of the eldest son = 3+7x = 3+28 = 31.</span>
The true value is 25.7 ml.
The calculated error is 15.6%.
Thus, the error amount equals 0.156 times 25.7, which calculates to 4.0092 ml.
The percentage error indicates that the student's measurement could either exceed or fall short of the true value by this error amount.
This leads to two potential readings:
one possibility is: 25.7 + 4.0092 = 29.7092 ml
the other possibility is: 25.7 - 4.0092 = 21.6908 ml
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 likelihood of selecting one girl is calculated as 
. This is based on having 5 girls within a total of 12 students, and the probability of an event can be expressed as: 
.
Using the same reasoning, for the next student, we have reduced the number of students by 1, leading to 11 possible outcomes instead of 12, giving us:
, which represents the probability of selecting a boy as the second choice.
Lastly, the probability of choosing a girl for the third selection follows the same logic and is given as:
.
However, we must combine these individual probabilities to determine the likelihood of this specific sequence of selections occurring:
This simplifies to:
To find the volume of an aluminum cylinder with a mass of 12.4 g and a length of 2.00 cm:
V = π r² L
Density of aluminum is 2.7 g / cm³
Therefore, V = 12.4 g: 2.7 g/cm³ ≈ 5 cm³
5 = 3.14 · r² · 2
r² = 5: 6.28
r² = 0.796
r = √0.796
Thus, the radius of the aluminum cylinder is 0.9 cm.
Answer:
The radius of an aluminum cylinder is 0.9 cm.
