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jonny [76]
2 years ago
15

Dan's school is selling tickets to a play. On

Mathematics
1 answer:
katrin [286]2 years ago
4 0

Answer: $9.50

Step-by-step explanation:Let's define the variables:

A = price of one adult ticket.

S = price of one student ticket.

We know that:

"On the first day of ticket sales the school sold 1 adult ticket and 6 student tickets for a total of $69."

1*A + 6*S = $69

"The school took in $150 on the second day by selling 7 adult tickets and student tickets"

7*A + 7*S = $150

Then we have a system of equations:

A + 6*S = $69

7*A + 7*S = $150.

To solve this, we should start by isolating one variable in one of the equations, let's isolate A in the first equation:

A = $69 - 6*S

Now let's replace this in the other equation:

7*($69 - 6*S) + 7*S = $150

Now we can solve this for S.

$483 - 42*S + 7*S = $150

$483 - 35*S = $150

$483 - $150 = 35*S

$333 = 35*S

$333/35 = S

$9.51 = S

That we could round to $9.50

That is the price of one student ticket.

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Aaron made a $400 purchase using his credit card. The credit card has a simple interest rate of 12% each year. How much will Aar
melisa1 [442]

Answer:  448 dollars

=============================================

Work Shown:

P = 400 is the balance

r = 0.12 is the decimal form of the interest rate

t = 1 is the number of years

Use those values to get the simple interest to be

i = P*r*t

i = 400*0.12*1

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5 0
3 years ago
Erica is a sheep farmer. She is having a problem with wolves attacking the flock. She starts with an initial 200 sheep and notic
Free_Kalibri [48]

The equation for the situation is y=200(\frac{2}{3})^{x}

It will take around 9 years for her to have around 5 sheep

Step-by-step explanation:

The exponential decay growth/decay equation is y=a(b)^{x} , where

  • a is the initial value
  • b is the growth/decay factor
  • If b > 1, then it is a growth factor
  • If 0 < b < 1, then it is a decay factor

Erica is a sheep farmer. She is having a problem with wolves attacking the flock. She starts with an initial 200 sheep and notices that the population is two-thirds of the previous year

∵ The population is decreased

∴ The equation is decay

∵ The population is two-thirds of the previous year

∴ The decay factor is \frac{2}{3}

∵ She starts with an initial 200 sheep

∴ the initial value is 200

∵ The decay equation is y=a(b)^{x} , where y represent the

   population in x years

∵ a = 200

∵ b = \frac{2}{3}

∴ y=200(\frac{2}{3})^{x}

The equation for the situation is y=200(\frac{2}{3})^{x}

∵ The population after x years is around 5 sheep

- Substitute y by 5 to find x

∵ 5=200(\frac{2}{3})^{x}

- Divide both sides by 200

∴ 0.025=(\frac{2}{3})^{x}

- Insert ㏒ for both sides

∴ log(0.025)=log(\frac{2}{3})^{x}

∴ log(0.025)=xlog(\frac{2}{3})

- Divide both sides by log(\frac{2}{3})

∴ 9.098 = x

∴ x is around 9 years

It will take around 9 years for her to have around 5 sheep

Learn more:

You can learn more about the equation  in brainly.com/question/10666510

#LearnwithBrainly

8 0
3 years ago
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