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rewona [7]
3 years ago
15

The bottom portion of a loading bin is cone shaped. The base radius of this part of the bin is 3.5 feet and the slant height is

6.5 feet. What is the capacity and lateral surface area of this part of the bin? Round your answer to the nearest hundredth. Lateral Surface Area = sq. ft. Volume = cu. ft.
Mathematics
2 answers:
jek_recluse [69]3 years ago
7 0

\bf \textit{lateral surface of a cone}\\\\ LA=\pi r\sqrt{r^2+h^2}~~ \begin{cases} ~~ r=radius\\ sh=\stackrel{slant~height}{\sqrt{r^2+h^2}}\\[-0.5em] \hrulefill\\ r=3.5\\ sh=6.5 \end{cases}\\\\\\ LA=\pi (3.5)(6.5)\implies LA\approx71.47 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{sh}{6.5}=\sqrt{r^2+h^2}\implies 6.5=\sqrt{3.5^2+h^2}\implies 6.5^2=3.5^2+h^2 \\\\\\ 6.5^2-3.5^2=h^2\implies \sqrt{6.5^2-3.5^2}=h\implies \sqrt{30}=h \\\\[-0.35em] ~\dotfill

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3.5\\ h=\sqrt{30} \end{cases}\implies V=\cfrac{\pi (3.5)^2\sqrt{30}}{3}\implies V\approx 70.26

blsea [12.9K]3 years ago
7 0

Answer to Q1:

A  = 71.47 sq.ft

Step-by-step explanation:

We have given the base radius and the slant height of the cone.

base radius  = r = 3.5 feet and slant height  = √r²+ h² = 6.5 feet

We have to find the lateral surface area of the cone.

The formula to find the lateral surface area of the cone:

A  = πr√r²+h²

Putting values in above formula, we have

A  = π(3.5)(6.3)

A  = 71.47 sq.ft which is the answer.

Answer to Q2:

V  = 70.26 cubic ft

Step-by-step explanation:

We have given the base radius and the slant height of the cone.

base radius  = r = 3.5 feet and slant height  =  √r²+h² = 6.5 feet

We have to find the volume of the cone.

The formula to find the lateral surface area of the cone:

V  = πr²h / 3

√r²+h² = 6.5

√3.5²+h²  = 6.5

h  = √30

Putting values in above formula, we have

V  = π(3.5)²(√30) / 3

V = π (12.25)√30 / 3

V  = 70.26 cubic ft which is the answer.

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Step-by-step explanation:

Correct statement of the question is ;

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<u>b) for balck ball</u>

P(b_{1} )= \frac{1}{3}

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P(w_{2} )=P(\frac{w_{2} }{w_{1} } ).P(w_{1} )+P(\frac{w_{2} }{b_{1} } ).P(b_{1} )

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Then the probability that white ball was transferred from urn 1 to urn 2 is;

P(\frac{white ball transfer }{w_{2} } )= \frac{P(\frac{w_{2} }{w_{1} } ).P(w_{2} )}{P(w_{2}) }

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Step-by-step explanation: You can find this answer by using the equation "14m + 24 = 16m + 14" and then solving for "m". To find what "m" is, you must first subtract 14m from both sides, leaving "24 = 2m + 14". Next, you subtract 14 from both sides, leaving "10 = 2m". Finally, you divide both sides by 2, which will give you "m = 5". Thus meaning each multiple choice question is worth 5 points.

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