Answers:
The first five terms are: 3, 8, 13, 18, 23
This is a linear sequence
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Explanation:
Plug in n = 1 to get the first term
E(n) = 5*n-2
E(1) = 5*1-2
E(1) = 3
So 3 is the first term
Do the same for n = 2 to get the second term
E(n) = 5*n-2
E(2) = 5*2-2
E(2) = 8
So 8 is the second term
And again for n = 3
E(n) = 5*n-2
E(3) = 5*3-2
E(3) = 13
This means 13 is the third term
And again for n = 4
E(n) = 5*n-2
E(4) = 5*4-2
E(4) = 18
This means 18 is the fourth term
Finally, repeat for n = 5
E(n) = 5*n-2
E(5) = 5*5 - 2
E(5) = 23
So 23 is the fifth term
This sequence is linear because we're incrementing by the same amount each time. In this case, we're adding 5 each time for the sequence 3, 8, 13, 18, 23, and so on.
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:
h = 1.4
c = 2.8
Step-by-step explanation:
For each problem, remember the special triangle side ratios then use a proportion. To solve, isolate the variable.
For the triangle with the variable h:
Since two of the angles are 45, this is an isosceles triangle. All isosceles triangles have two equal sides that are not the hypotenuse.
In a right isosceles triangle, the ratio for regular side to hypotenuse is 1 to √2.
h = √2
h ≈ 1.4
For the triangle with the variable c:
The is an equilateral triangle cut in half because the angles are 30 and 60.
The side ratio of altitude to hypotenuse is √3 to 2.
c = 2√2
c ≈ 2.8
