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erma4kov [3.2K]
3 years ago
7

4. What is the median of this data set?

Mathematics
2 answers:
Stolb23 [73]3 years ago
8 0

Answer:

5.5

Step-by-step explanation:

because the middle most number is 5,6

so

5+6=11

11/2= 5.5

Dima020 [189]3 years ago
8 0

Answer : The median of the given set of data of values is, 5.5

Step-by-step explanation :

Median : It is the middle term that is sorted by the list of numbers. It is determine by placing the numbers in increasing order.

For odd observations the formula will be: \frac{n+1}{2}

For even observations the formula will be: \frac{\text{Average of two middle value}}{2}

As we are given that the set of data:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Now we have to determine the median.

For even data:

\frac{\text{Average of two middle value}}{2}

Middle two values are, 5 and 6

\frac{5+6}{2}=5.5

Thus, the median of the given set of data of values is, 5.5

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Step-by-step explanation:

First, converting R percent to r a decimal

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A 180 ft long board is cut into 3 pieces. The second piece of
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Answer:

7. Approximately 402.7 inches or 2pi times 64

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Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

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4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

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For 7-8, divide through each term by the largest power of x in the expression:

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8. Divide through by x² again:

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9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

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10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

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