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3 years ago

The cost to attend a Wildcat sports event is: Adult: $7 Student: $5 The total sales was $700.

Mathematics

1 answer:

hram777 [196]3 years ago

4 0

Answer:

7x+5y=700

91 student tickets

Step-by-step explanation:

7x+5y=700

7(35)+5y=700

245+5y=700

5y=455

y=91

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How many ways can a person select 3 coins from a box consisting of a penny, a nickel, a dime, a quarter, a half dollar, and a on

tekilochka [14]

Answer:

A person can select 3 coins from a box containing 6 different coins in 120 different ways.

Step-by-step explanation:

Total choices = n = 6

no. of selections to be made = r = 3

The order of selection of coins matter so we will use permutation here.

Using the formula of Permutation:

nPr = $\frac{n!}{(n-r)!}$

We can find all possible ways arranging 'r' number of objects from a given 'n' number of choices.

Order of coin is important means that if we select 3 coins in these two orders:

--> nickel - dime - quarter

--> dime - quarter - nickel

They will count as two different cases.

Calculating the no. of ways 3 coins can be selected from 6 coins.

nPr = $\frac{n!}{(n-r)!}$ = $\frac{6!}{(6-3)!}$

nPr = 120

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3 years ago

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3 years ago

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Nana76 [90]

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Step-by-step explanation:

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2 years ago

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3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

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3 years ago