Help? which one should I choose

34,000 people attended a ballgame at a stadium that offers two kind of seats: general admission and reserved. The day's receipts

hoa [83]

Answer:

The number of people who paid $ 12 for reserved seat is 5,000

The number of people who paid $ 4 for general seat is 29,000

Step-by-step explanation:

Given as :

The total number of people attending a ballgame = 34,000

The total receipt of the ticket's seat = $ 176,000

The amount pad for reserved seat = $ 12

The amount paid for general admission = $ 4

Let The number of people for reserved seat = r

And The number of people for general admission = g

Now, According to question

The total number of people attending a ballgame = The number of people for reserved seat + The number of people for general admission

or, r + g = 34,000 ...........1

The total receipt of the ticket's seat = The amount pad for reserved seat × The number of people for reserved seat + The amount paid for general admission × The number of people for general admission

Or, $ 12 × r + $ ×4 g = $ 176,000 .........2

or, $ 12 × ( r + g ) = $ 12 × 34000

Or, $ 12 r + $ 12 g = $ 408,000

Solving equation

( $ 12 r + $ 12 g ) - ($ 12 r + $ 4 g ) = $ 408,000 - $ 176,000

Or, ( $ 12 r - $ 12 r ) + ( $ 12 g - $ 4 g ) = $ 232,000

Or 0 + 8 g = 232,000

∴ g = $\frac{232000}{8}$

I.e g = 29,000

So , The number of people for general admission = g = 29,000

Put the value of g in Eq 1

I.e r + g = 34,000

or , r = 34,000 - g

∴ r = 34000 - 29000

I.e r = 5,000

So, The number of people for reserved seat = r = 5,000

Hence The number of people who paid $ 12 for reserved seat is 5,000

And The number of people who paid $ 4 for general seat is 29,000 Answer

6 0

3 years ago

What is the volume of the shape?

rodikova [14]

Answer:

volume: 1/3 a²h

= 1/3 ×2²x6

= 8 feet³

6 0

3 years ago

Increase £25 by 25% . .

Stells [14]

£31.25 homie yuh yuh

8 0

3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago

How many solutions does the equation 4x + 8 = − 2 − 2 + 5x have?

ikadub [295]

Answer:

12 =x

There is one solution

Step-by-step explanation:

4x + 8 = − 2 − 2 + 5x

Combine like terms

4x+8 = -4+5x

Subtract 4x from each side

4x-4x+8 = -4+5x-4x

8 = -4+x

Add 4 to each side

8+4 = -4+4+x

12 =x

6 0

3 years ago

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