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grin007 [14]
3 years ago
14

2. The school PTA is sponsoring a dance. They decide to price student tickets at $6.00. It will cost the PTA $250.00 to put on t

he dance. How many tickets will the PTA need to sell in order to make a profit?​
Mathematics
2 answers:
weqwewe [10]3 years ago
8 0

Answer:

The PTA must sell 42 tickets to make a profit.

Step-by-step explanation:

To solve this problem, we will have to construct an inequality in which $250 is less than our solution.

Let s = student tickets

Since each student ticket is priced at $6.00, the money raised by the student tickets can be represented by the expression:

6s

Since it will cost the PTA $250.00 for the dance, the profit will have to be equal to anything greater than $250.00.

Set up your inequality:

250

Divide both sides of the inequality by the coefficient of s, which is 6:

41.7=s

As you can see, we are left with a decimal, but we will have to round to the nearest whole number as the tickets can only be represented as whole numbers:

42=s

Therefore, the PTA must sell 42 tickets in order to make a profit.

-

You can check your work by substituting the solved value for the tickets into the inequality:

250

250

250

And since $252 is greater than $250 (represents a $2 profit), our solution is correct.

Tatiana [17]3 years ago
3 0

Answer:

42 tickets.

Step-by-step explanation:

To make a profit, the PTA will need to earn at least $251, because it spent $250 on the dance. so we can divide 251/6 = 41 remainder 5, which means we need to sell 42 tickets.

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The value of x is -2

Step-by-step explanation:

Given that

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Expand the bracket                    

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Collect like terms

12x - 3x = 6 - 6 - 18

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Divide through by 9

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Learn more about inequality here : brainly.com/question/20252161

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riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

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\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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