Height = x (This is given)
Length = 5x
Width = 6x
Volume = length × width × height
The equation will be: (5x)(6x)(x) = 1920
simplify:
30x² × (x) = 1920
30x³ = 1920
Divide both sides by 30
x³= 64
x = ∛64
x= 4
If x = height, then, the height is 4cm
5x = length
20cm = length
6x = width
24cm = width
The first one is 200 cause 360-160=200 cause it is the arc
Answer:
a) 1/64
b) 1/4096
Step-by-step explanation:
As you can tell from the example, the exponent of 1/2 is the number of heads in a row.
a) p(6 heads in a row) = (1/2)^6 = 1/(2^6) = 1/64
b) p(12 heads in a row) = (1/2)^12 = 1/(2^12) = 1/4096
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<em>Additional comment</em>
The probability of a head is 1/2 because we generally are concerned with a "fair coin." That is defined as a coin in which each of the 2 possible outcomes has the same probability, 1/2. Similarly, a "fair number cube" has 6 faces, and the probability of each is defined to be the same as any other, 1/6. Loaded dice and unfair coins do sometimes show up in probability problems.
8. The mean of the grades (74) is substantially below the median (88), so the distribution is not symmetrical. Of course, half the grades are below the median, just as half are above the median. However, the mean being 88-74 = 14 points below the median means that the lower half of the grades will have an average at least twice that much, or 28 points, below the median. The distribution of grades must extend quite a bit further to the left of the median than it does to the right. Hence ...
B. The distribution is skewed left.
9. It seems likely the distribution has a number of low grades pulling the average down. There certainly exists the possibility that at least one of them qualifies as an outlier by being more than 1.5 times the IQR below the first quartile. That rule, however, only applies when the distribution is relatively symmetrical, which this one is not. There does not appear to be any recommended way to describe an outlier when the distribution is skewed and has a long tail.*
TRUE: The distribution likely has an outlier.
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* Math instruction these days rarely recognizes such subtleties. Since it is probable that at least one value is well below the bottom quartile, I've shown my guess at the expected answer as TRUE. In a real set of grades, I expect the tail of the distribution to have enough low grades that they cannot be considered to be outliers.
Answer:
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Step-by-step explanation:
Solution: 15/2 = 7.5
7/2 = 3.5
The smaller number = 7.5-3.5 or 4, and the larger number is 7.5+3.5 or 11. Answer.
