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weeeeeb [17]
2 years ago
15

Just the first one please ss soon as possible

Mathematics
1 answer:
12345 [234]2 years ago
4 0

Answer:

Step-by-step explanation:

zamn

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#89 will give brainliest for best answer!​
Lady_Fox [76]

Answer:

Median: 11

Range: 12

25th Percentile: 9

75th Percentile: 14

Interquartile Range: 5

Step-by-step explanation:

The median is the line in the 'middle' of the box. The range is the largest number minus the smallest number. 25th Percentile is the first line of the box, 75th is 3rd line. IQR is the 75th Percentile - 25th Percentile.

Hope this helps! :)

6 0
2 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=
2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

6 0
3 years ago
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