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Westkost [7]
3 years ago
14

What is 682.717 rounded to the nearest tenth?

Mathematics
2 answers:
GuDViN [60]3 years ago
6 0

Answer:

682.7

Step-by-step explanation:

because you cant round it up

Alla [95]3 years ago
4 0

Answer:

THe answer is eather 682.72 or 680 but you have to be more porsfic

Step-by-step explanation:

First i go to 8 in 682.717 and se is the number befor is

if it 5 or more we add 1 more but if its 4 or less we let it rest

so we let about 2 is bothow so now it gose to 680 and first i go to the 1 in the number then look back and it is above so i add one more so now you do add one more and the reast tearn 0's.Now this is the answer 682.72

MY answer 680 i think is the right answer

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Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

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3 years ago
Anthony bought 3 chicken wings for $3.00. What's the unit cost of one wing?
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Answer; $1
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2 years ago
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OLga [1]

Answer:

Answers may vary but will most likely be close to 2.

Step-by-step explanation

  1.        Given:

                 first test:38%

                 second test:76%

    SIMULATION FIRST TEST

Randomly select a 2-digit number.

If the digit is between 00 and 35 then you passed the test,else you did not pass the test.

SIMULATION SECOND TEST

Randomly select a 2-digit number.

if the digit is between 00 and 75 then you passed the test,else you did not pass the test.

SIMULATION TRIAL

Perform the simulation of the first test.if you did not pass the first test then perform the simulation of the second test.

Record the number of trials needed to pass the first or second test.

Repeat 20 times and take the average of the 20 recorded number of trials

(what is the sum of recorded values divided by 20).

Note:you will most likely obtain a result of about two trials needed.

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kvv77 [185]

Answer:

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Step-by-step explanation:

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3 years ago
What is the answer 8x-22=-60
solmaris [256]

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7 0
3 years ago
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