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V125BC [204]
3 years ago
5

A theater has a seating capacity of 750 and charges $3 for children, s5 for students, and $7 for adults. At a certain screening

with fll ttendance, there were combined. The receipts totaled $3450. How many children attended the show?
Mathematics
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

Between 150 and 450

Step-by-step explanation:

We are going to find the number by resolving  a system of linear equations.

First we write the system equations :

C+S+A=750

Where C : children, S : students and A : adults

The equation represents the ''full attendance''

The second equation :

3C+5S+7A=3450

This equation represents the totaled receipts.

The system :

C+S+A=750\\3C+5S+7A=3450

has the following associated matrix :

\left[\begin{array}{cccc}1&1&1&750\\3&5&7&3450\end{array}\right]

By performing elementary matrix operations we find that the matrix is equivalent to

\left[\begin{array}{cccc}1&0&-1&150\\0&1&2&600\\\end{array}\right]

The new system :

C-A=150\\S+2A=600

Working with the equations :

C = 150 + A\\S = 600-2A

Our solution vector is :

\left[\begin{array}{c}C&S&A\end{array}\right] =\left[\begin{array}{c}150+A&600-2A&A\end{array}\right]

For example :

If 0 adults attended ⇒ A = 0

C = 150 + 0 \\C = 150\\S = 600 - 2A\\S = 600

This verify the totaled receipts equation :

150($3)+600($5) = $ 3450

A ≥ 0 ⇒ If A = 0 ⇒ C = 150

C = 150 is the minimum children attendance

From the equation :

S = 600 -2A

S ≥0

600 - 2A ≥ 0

600 ≥ 2A

300≥ A

A is restricted to the interval [ 0, 300]

When A = 0 ⇒ C = 150

When A = 300 ⇒C = 150 + A = 150 + 300 = 450

Children ∈ [ 150,450]

With C being an integer number (including 0)

Also S and A are integer numbers (including 0)

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MCR3U1 Culminating 2021.pdf
11111nata11111 [884]

Answer:

(a) y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is 607,325

(ii) The population after 24 hours is 1,828,643

(c) The rate of increase of the population as a percentage per hour is 7.132%

(d) The doubling time of the population is approximately, 10.06 hours

Step-by-step explanation:

(a) The initial population of the bacteria, y₁ = a = 350,000

The time the colony grows, t = 12 hours

The final population of bacteria in the colony, y₂ = 800,000

The exponential growth model, can be written as follows;

y = a \cdot (1 + r)^t

Plugging in the values, we get;

800,000 = 350,000 \times (1 + r)^{12}

Therefore;

(1 + r)¹² = 800,000/350,000 = 16/7

12·㏑(1 + r) = ㏑(16/7)

㏑(1 + r) = (㏑(16/7))/12

r = e^((㏑(16/7))/12) - 1 ≈ 0.07132

The  model is therefore;

y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is given as follows;

y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82

By rounding down, we have;

The population after 8 hours, y = 607,325

(ii) The population after 24 hours is given as follows;

y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571

By rounding down, we have;

The population after 24 hours, y = 1,828,643

(c) The rate of increase of the population as a percentage per hour =  r × 100

∴   The rate of increase of the population as a percentage = 0.07132 × 100 = 7.132%

(d) The doubling time of the population is the time it takes the population to double, which is given as follows;

Initial population = y

Final population = 2·y

The doubling time of the population is therefore;

2 \cdot y = y \times (1 + 0.07132)^t

Therefore, we have;

2·y/y =2 = (1 + 0.07132)^t

t = ln2/(ln(1 + 0.07132)) ≈ 10.06

The doubling time of the population is approximately, 10.06 hours.

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