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tangare [24]
3 years ago
8

The cooking club made some pies to sell at a basketball game to raise money for the new math books. The cafeteria contributed tw

o pies to the sale. Each pie was then cut into six pieces and sold.There were a total of 60 pieces to sell. How many pies did the club sell
Mathematics
1 answer:
Neko [114]3 years ago
3 0

Answer:

10

Step-by-step explanation:

Each pie was cut into 6 pieces. 6 x 10= 60.

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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
You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e
victus00 [196]

You intend to estimate a population mean μ from the following sample. 26.2 27.7 8.6 3.8 11.6 You believe the population is normally distributed. Find the 80% confidence interval.  Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places.

Answer:

The  Confidence interval = (8.98 , 22.18)

Step-by-step explanation:

From the given information:

mean = \dfrac{ 26.2+ 27.7+ 8.6+ 3.8 +11.6 }{5}

mean = 15.58

the standard deviation \sigma = \sqrt{\dfrac{\sum(x_i - \mu)^2 }{n}}

the standard deviation = \sqrt{\dfrac{(26.2  - 15.58)^2 +(27.7 - 15.58)^2 +(8.6  - 15.58)^2 + (3.8  - 15.58)^2  + (11.6  - 15.58)^2  }{5 }  }

standard deviation = 9.62297

Degrees of freedom df = n-1

Degrees of freedom df = 5 - 1

Degrees of freedom df = 4

For df  at 4 and 80% confidence level, the critical value t from t table  = 1.533

The Margin of Error M.O.E = t \times \dfrac{\sigma}{\sqrt{n}}

The Margin of Error M.O.E = 1.533 \times \dfrac{9.62297}{\sqrt{5}}

The Margin of Error M.O.E = 1.533 \times 4.3035

The Margin of Error M.O.E = 6.60

The  Confidence interval = ( \mu  \pm M.O.E )

The  Confidence interval = ( \mu  +  M.O.E , \mu - M.O.E )

The  Confidence interval = ( 15.58 - 6.60 , 15.58 + 6.60)

The  Confidence interval = (8.98 , 22.18)

7 0
3 years ago
Every 3 hours, a machine produces 60 baskets what is the unit rate
uranmaximum [27]

Answer:

20 per hour

Step-by-step explanation:

divide 60 by 3=20

8 0
3 years ago
Read 2 more answers
At the fast pack shiping. The employees can unload 25 trucks in 5 hours which of the following is a correct unit rate for this s
Natasha2012 [34]

Answer:

5 trucks per hour

Step-by-step explanation:

25 trucks ÷ 5

5 hours ÷ 5

= 5 trucks per hour

3 0
3 years ago
A tank contains 3,000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 30 L/min. The solution is
Shalnov [3]

Answer:

Step-by-step explanation:

Volume of tank is 3000L.

Mass of salt is 15kg

Input rate of water is 30L/min

dV/dt=30L/min

Let y(t) be the amount of salt at any time

Then,

dy/dt = input rate - output rate.

The input rate is zero since only water is added and not salt solution

Now, output rate.

Concentrate on of the salt in the tank at any time (t) is given as

Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000

dy/dt= dV/dt × dM/dV

dy/dt=30×y/3000

dy/dt=y/100

Applying variable separation to solve the ODE

1/y dy=0.01dt

Integrate both side

∫ 1/y dy = ∫ 0.01dt

In(y)= 0.01t + A, .A is constant

Take exponential of both side

y=exp(0.01t+A)

y=exp(0.01t)exp(A)

exp(A) is another constant let say C

y(t)=Cexp(0.01t)

The initial condition given

At t=0 y=15kg

15=Cexp(0)

Therefore, C=15

Then, the solution becomes

y(t) = 15exp(0.01t)

At any time that is the mass.

5 0
3 years ago
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