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enyata [817]
3 years ago
8

It costs 135.25 for 15 child tickets ti the museum.Adult tickets cost 12.95 each. Find the cost for 40 children to attend along

with 5 adult chaperones
Mathematics
1 answer:
Masja [62]3 years ago
4 0

Answer:the cost for 40 children to attend along with 5 adult tickets = $ 417.55

Step-by-step explanation:

It costs 135.25 for 15 child tickets to the museum. Let us determine the cost per child ticket to the museum.

cost per child ticket to the museum.

= total cost of 15 child tickets / number of child tickets.

= 132.25 / 15 = 8.82

Adult tickets cost 12.95 each.

The cost of 40 child tickets is

40 × 8.82 = $352.8

The cost of 5 adult tickets = 5 × 12.95

=$ 64.75

the cost for 40 children to attend along with 5 adult tickets would be

352.8 + 64.75 = $ 417.55

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Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
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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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Suppose that you earned a​ bachelor's degree and now​ you're teaching high school. The school district offers teachers the oppor
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Answer:

Step-by-step explanation:

Answer:

a. The amount that is saved at the expiration of the 5 year period is $22,769.20¢

b. The amount of interest is $2,769.20¢

Step-by-step explanation:

Since the amount that is deposited every year for a period of five years is $4,000 and the rate of the interest is 6.5%. We can always calculate the amount that is saved at the expiration of the five years.

    We will first state the formula for calculating the future value of annuity:-

      Future value of annuity =

                      P[\frac{(1 + r)^{t}-1 }{r}]

   Where P is the amount deposited per year.

   r is the rate of interest

   t is the time or period

 

    and in this case, the actual value of P = $4,000

      rate of interest, r is 6.5% = 0.065

      time, t is 5 years.

   Substituting e, we have:

   Fv of annuity =

                          4,000[\frac{(1 + 0.065)^{5}-1 }{0.065 }]

   = 4,000 × [((1.065)^5)- 1/0.065]

 = 4,000 × [(1.37 - 1)/0.065]

 = 4,000 × (0.37/0.065)

 = 4,000 × 5.6923

 = $22,769.20¢

a. Therefore the amount that is saved at the end of the five (5) years is $22,769.20¢

b. To find the interest, we will calculate the amount of deposit made during the period of five years and subtract the sum from the current amount that is saved ($22,769.29¢).

  Since I deposited 4,000 every year for five years, the total amount of deposit I made at the period =

       4,000 × 5 = $20,000

  The amount of interest is then = $22,769.20¢ - $20,000 = $2,769.20¢

3 0
3 years ago
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