Can someone help me out with this problem

Mathematics

1 answer:

Nostrana [21]3 years ago

8 0

I need more information.

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What is the volume of a hemisphere with a radius of 3.1 in, rounded to the nearest tenth of a cubic inch?

bogdanovich [222]

<h3>Answer:</h3>

<u>Volume of hemisphere = 2/3πr³</u>

2/3 × 3.14 × 3.1³
187.08/3
62.36 or 62.4 rounded.

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3 years ago

For number 6, evaluate the definite integral.

maks197457 [2]

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$

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3 years ago

Here is the histogram of a data distribution. All class widths are 1.

prisoha [69]

Answer:

Step-by-step explanation:

The median is the middle value in the data distribution.

From 1 → 4 there are14 values ( count the blocks )

From 5 → 10 there are 14 values

Thus the middle value of the distribution is 5, that is

median is 5 → D

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