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2 years ago

This box plot shows the heights ( in feet) of a sample of pine trees

Mathematics

2 answers:

kozerog [31]2 years ago

6 0

Answer:

Step-by-step explanation:

The new one would have a long gap between the Q3 and the maximum of the box and whisker plot, therefore it should create a positive skew.

Lady_Fox [76]2 years ago

3 0

Answer:

Step-by-step explanation:

It will be positively skewed (skewed right), because that means that the longer segment would be on the right side of the median.

140 is a major outlier and a lot larger than the more central numbers. However, it's only 1 number, so the median (location of the "box") would remain similar or the same. The end on the right, however, would move far to the right, extending the "line" on that side.

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mario62 [17]

B, that is where the line crosses the x axis

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2 years ago

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aalyn [17]

Answer: 3 3/5

3 3/4=3.75

3 3/5=3.6

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2 years ago

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

On integration , we get

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$

After simplification and cancellation we get

= $\frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6$

on calculation , we get

= $\frac{124,416}{5} + \frac{864}{3} + 180$

On L.C.M 15

= $\frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}$

= 25 351.2 units²

Conclusion:-

The area of the region between the graph of the given function and the x-axis = 25,351 units²

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