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3 years ago

10

BRAINLIST PLEASEEEE HELP

2 answers:

Aleksandr [31]3 years ago

6 0

Answer:

About 527.79

Step-by-step explanation:

formula is A=2πrh+2πr^2

arsen [322]3 years ago

5 0

Answer:

$527.52\:\mathrm{in^2}$

Step-by-step explanation:

The surface area of this cylinder consist of two circles and a 3D outer edge with respect to the circumference. The area of a circle is given as $r^2\pi$ . Using $\pi=3.14$ as requested in the problem, the area of both these circles is:

$2\cdot6^2\cdot 3.14=226.08\:\mathrm{in^2}$ .

The area of the outer edge is the circumference of the circle multiplied by the height. The circumference of a circle is given by $2r\pi$ .

Therefore, the circumference's 3D outer edge is:

$8\cdot2\cdot6\cdot 3.14 = 301.44\:\mathrm{in^2}$ .

Adding these up, the total surface area is:

$226.08+301.44=\fbox{$527.52\:\mathrm{in^2}$}$ .

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ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these th

Misha Larkins [42]

Answer:

a. $P(E_1/A)=0.0789$

b. $P(E_2/A)=0.395$ \

c. $P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

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$P(A/E_1)=0.01$

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$P(A/E_3)=0.10$

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem, $P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}$

We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$

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4 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

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