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3 years ago

The volume of a sphere is 45001 m3. Find the length of the radius of the great circle.

Mathematics

1 answer:

Scilla [17]3 years ago

6 0

Answer:

$\large \boxed{\text{22.07 m}}$

Step-by-step explanation:

The formula for the volume (V) of a sphere is:

V = ⁴/₃πr³

We can use this equation to calculate the radius (r).

$\begin{array}{rcl}45001 & = & \dfrac{4}{3} \pi r^{3}\\\\33750 & = &\pi r^{3}\\10740 & = &r^{3}\\r & = & \textbf{22.07 m}\\\end{array}\\\text{The radius of the great circle is $\large \boxed{\textbf{22.07 m}}$}$

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Divide. 2/7 divided by 1/3

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3 years ago

The lengths of a particular snake are approximately normally distributed with a given mean mu = 15 in. and standard deviation si

Helga [31]

The percentage of the snakes is longer than 16.6 in. with a mean of 15 in. and a standard deviation of 0.8 in. is 2.275%.

<h3>What is a normal distribution?</h3>

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

The lengths of a particular snake are approximately normally distributed with a given mean = 15 in. and standard deviation = 0.8 in.

Then the percentage of the snakes is longer than 16.6 in. will be

$z = \dfrac{ x - \mu }{ \sigma }\\\\z = \dfrac{16.6-15}{0.8}\\\\z = 2$

Then we have

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More about the normal distribution link is given below.

brainly.com/question/12421652

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2 years ago

Finish

alex41 [277]

Answer:

What is this? Is that a question? I mean it doesn't seem like one. can you post the proper one?

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3 years ago

Read 2 more answers

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

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3 years ago