Ron and kathy are ticket sellers at their class play, ron handling students tickets that sell for $2.00 each and kathy selling adult tickets for $4.50 each. if their total income for 364 tickets was $1175.50, how many did ron sell?

Let's create two equations and solve them using the substitution method.

where s = student tickets and a = adult tickets.

Solve the first equation for s.

Subtract a from both sides.s = 364 - a

Plug s into the second equation.

2(364 - a) + 4.5a = 1175.5Distribute 2 inside the parentheses.728 - 2a + 4.5a = 1175.5Combine like terms.728 + 2.5a = 1175.5Subtract 728 from both sides.2.5a = 447.5Divide both sides by 2.5.a = 179

Plug 179 for a into the first equation.

s + 179 = 364Subtract 179 from both sides.s = 185

Since Ron handled student tickets, he sold 185 tickets (s = student tickets = 185).

Let's check our work.

In the first equation, s + a = 364; let's plug in what we got for a and s into the equation.

This means we will plug in 179 for a and 185 for s.

185 + 179 = 364

189 + 179 does equal 364, so this means we are correct in solving this system of equations.

Hope this helped you^

Part A:

Linear function

Part B:

House 1; f(x) = 7595.3·x + 178515

House 2; f(x) = 11,000·x + 179000

Part C:

Yes, there will be a significance difference

The value of house 1 after 30 years is $406,374

The value of house 2 after 30 years is $509,000

Step-by-step explanation:

The given values are

               House 1

Year 1,    186,160

Year 2,   193606.40

Year 3,   201350.66

We see that the prices of the houses have a common difference given as follows;

201,350.66 - 193606.40 = 7744.26,

193,606.40 - 186,160 = 7446.4

Which is fairly constant giving a linear function of the house after a fixed number of years

               House 2

Year 1,    190,000

Year 2,   201,000

Year 3,   212,000

We see that the prices of the houses have a common difference given as follows;

212,000 - 201,000 = 11,000,

201,000 - 190,000 = 11,000

Which is fairly constant

Which is fairly constant giving a linear function of the house after a fixed number of years

Part B:

The function for the first house  is given by the regression formula from regression analysis

Y = a + b·X

Where:

X = Value of the house at Year Y

Y = The year

Which gives;

f(x) = 7595.3·x + 178515

The function for the first house  is given by the arithmetic progression also by regression analysis as follows;

f(x) = 11,000·x + 179000

Part C

The value of house 1 after 30 years = 7595.3×30 + 178515 = $406,374

The value of house 2 after 30 years = 11,000×30 + 179000 = $509,000

Yes, there will be a significance difference between house 2 and house 1 to about 509,000 - 406,374 = $102,626.