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3 years ago

A can of pineapple juice is a cylinder with a radius of 2.4in.and a height of 11 in. what is the area of the label around the ca

n? PLEASE HELP

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

4 0

$\bf \textit{lateral surface area of a cylinder}\\\\
A=2\pi r h\qquad \begin{cases}
r=radius\\
h=height\\
-----\\
r=2.4\\
h=11
\end{cases}\implies A=2\pi (2.4)(11)\implies A=52.8\pi$

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3 years ago

Determine whether the sequences converge.

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$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

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The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

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Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

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Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

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Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

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3 years ago

In the game of roulette, a player can place a $9 bet on the number 11, and have a 1/38 probability of winning. If the metal ba

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Answer:

1. What is the expected value of the game to the player?

Expected value= -$0.47368

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Expected value= -$473.68

Step-by-step explanation:

1. What is the expected value of the game to the player?

Expected value can tell how much you will gain/lose every time you do the game. To find the expected value you need to multiply every value with its chance to occur. There are two types of events here, winning and losing. You have 1/38 chance of winning $315 and 37/38 chance of losing $9.

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You are expected to lose $0.47368 every time you play

2. If you played the game 1000 times, how much would you expect to lose?

Expected value calculated for each roll. If you want to know how much you gain/lose by multiple rolls, you just need to multiply the expected value with the number of rolls. The game has negative expected value so you are expected to lose money instead of gaining money. The money you expected to lose will be:

number of roll * expected value

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You are expected to lose $473.68 if you play the game 1000 times.

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3 years ago

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Answer:

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2 years ago

Read 2 more answers

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Step-by-step explanation:

See attachment.

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10000 * ( 1 + 0 - {25/10000} )

10000 * ( 1 - {25/10000} )

10000 - {25*10000/10000}

10000 - {25}

9975

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3 years ago