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valentina_108 [34]
2 years ago
6

The owner of a movie theater is counting the receipts from one day of ticket sales. He knows that 250 tickets were sold. Adult t

ickets are $10 each and children’s tickets are $8 each. If the total collected for the day was $2276, how many of each type of ticket were sold? a. Write the system of equations to represent this situation. b. Solve the system using elimination. c. How many of each type of ticket were sold?
Mathematics
1 answer:
irinina [24]2 years ago
8 0

Step-by-step explanation:

x + y = 250 -eq.1

10x + 8y = 2276 -eq.2

eq.1 × 8

8x + 8y = 2000 -eq.3

eq.2 - eq.3

10x + 8y = 2276

( - ) 8x + 8y = 2000

2x = 276

x = 138

sub x = 138 into eq.1

138 + y = 250

y = 250 - 138

y = 112

x = 138, y = 112

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Using linked lists or a resizing array; develop a weighted quick-union implementation that removes the restriction on needing th
balu736 [363]

Answer:

Step-by-step explanation:

package net.qiguang.algorithms.C1_Fundamentals.S5_CaseStudyUnionFind;

import java.util.Random;

/**

* 1.5.20 Dynamic growth.

* Using linked lists or a resizing array, develop a weighted quick-union implementation that

* removes the restriction on needing the number of objects ahead of time. Add a method newSite()

* to the API, which returns an int identifier

*/

public class Exercise_1_5_20 {

public static class WeightedQuickUnionUF {

private int[] parent; // parent[i] = parent of i

private int[] size; // size[i] = number of sites in subtree rooted at i

private int count; // number of components

int N; // number of items

public WeightedQuickUnionUF() {

N = 0;

count = 0;

parent = new int[4];

size = new int[4];

}

private void resize(int n) {

int[] parentCopy = new int[n];

int[] sizeCopy = new int[n];

for (int i = 0; i < count; i++) {

parentCopy[i] = parent[i];

sizeCopy[i] = size[i];

}

parent = parentCopy;

size = sizeCopy;

}

public int newSite() {

N++;

if (N == parent.length) resize(N * 2);

parent[N - 1] = N - 1;

size[N - 1] = 1;

return N - 1;

}

public int count() {

return count;

}

public int find(int p) {

// Now with path compression

validate(p);

int root = p;

while (root != parent[root]) {

root = parent[root];

}

while (p != root) {

int next = parent[p];

parent[p] = root;

p = next;

}

return p;

}

// validate that p is a valid index

private void validate(int p) {

if (p < 0 || p >= N) {

throw new IndexOutOfBoundsException("index " + p + " is not between 0 and " + (N - 1));

}

}

public boolean connected(int p, int q) {

return find(p) == find(q);

}

public void union(int p, int q) {

int rootP = find(p);

int rootQ = find(q);

if (rootP == rootQ) {

return;

}

// make smaller root point to larger one

if (size[rootP] < size[rootQ]) {

parent[rootP] = rootQ;

size[rootQ] += size[rootP];

} else {

parent[rootQ] = rootP;

size[rootP] += size[rootQ];

}

count--;

}

}

public static void main(String[] args) {

WeightedQuickUnionUF uf = new WeightedQuickUnionUF();

Random r = new Random();

for (int i = 0; i < 20; i++) {

System.out.printf("\n%2d", uf.newSite());

int p = r.nextInt(i+1);

int q = r.nextInt(i+1);

if (uf.connected(p, q)) continue;

uf.union(p, q);

System.out.printf("%5d-%d", p, q);

uf.union(r.nextInt(i+1), r.nextInt(i+1));

}

}

}

8 0
3 years ago
The revenue, cost, and profit functions for a line of cell phone cases is shown. Identify the location on the profit function wh
grigory [225]

Answer:

Im pretty sure the profit location is at the center of the rise

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
a line passes through point (-2,5) and has a slope of 2/3 points a (x,3) and b ( -2, y) lie on the line what is the value of y a
Tomtit [17]

Answer:

The value of y is 5 , and The value of x is - 5

Step-by-step explanation:

Given as :

The line passes through point ( - 2, 5)

The slope of line is m =  \frac{2}{3}

Equation of line is written as

y - y_1 = m × (x - x_1)

Since line passes through line ( - 2 , 5 ) and have slope \frac{2}{3}

So, y - 5 =   \frac{2}{3}  × (x + 2)

Or, 3 × ( y - 5 ) = 2 × ( x + 2 )

Or, 3 y - 15 = 2 x + 4

Or, 3 y - 2 x - 19 = 0 is the equation of line

Now points ( x , 3 ) lies on the line 3 y - 2 x - 19 = 0

So, 3 × 3 - 2 x - 19 = 0

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I.e x = - 5

Again points ( - 2 , y )  lies on the line 3 y - 2 x - 19 = 0

So, 3 × y + 2 × 2 - 19 = 0

Or,  3 × y + 4 -19 = 0

or, 3 y = 15

∴  y = \frac{15}{3}

I.e y = 5

Hence The value of y is 5 , and The value of x is - 5   .  Answer

8 0
2 years ago
Determine how many real square root of 49 has.explain. please help me for explain
CaHeK987 [17]
49 only has one square root, which is 7. There is no other number that you can multiply by itself to get 49
7 0
3 years ago
Read 2 more answers
12n^2+48n = -n^3-64 factoring polynomial equations
xxTIMURxx [149]

Answer:

Step-by-step explanation:

12n2+48n=−n3−64

Step 1: Subtract -n^3-64 from both sides.

12n2+48n−(−n3−64)=−n3−64−(−n3−64)

n3+12n2+48n+64=0

Step 2: Factor left side of equation.

(n+4)(n+4)(n+4)=0

Step 3: Set factors equal to 0.

n+4=0

n=−4  

4 0
3 years ago
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