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LekaFEV [45]
3 years ago
13

The function f(x)=501170(0.98)x gives the population of a Texas City x years after 1995. What was the population in 1995?

Mathematics
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

The population of a Texas City in 1,995 was 501,170\ people

Step-by-step explanation:

we have

f(x)=501,170(0.98^{x})

we know that

The population in 1,995 is for x=0

so

substitute the value of x=0 in the function f(x)

f(0)=501,170(0.98^{0})=501,170\ people

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{0.23,19%1/5} Order each set of numbers from least to greatest show work
Liono4ka [1.6K]
First you have to change them all to the same form, so you have to pick one out of them, so for instance you could pick percent which would be; 23%, 19%, and 20%, so from least to greatest it would be; 19%, 20%, and 23%. Hope this helps! 
4 0
3 years ago
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
Which of the following is unbalanced?
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Answer:

A. 2Fe + 3CI2 - 2FeCI3 ........

5 0
2 years ago
Write a recursive definition for the sequence 14, 10, 6, 2,… Common difference is 4 what do I do next?
leonid [27]
A[1] = 14 . . . . . . . . . defines the first term
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6 0
3 years ago
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jan, mya, and Sara ran a total of 64 miles last week. jan and mya ran the same number of miles. Sara ran 8 less miles than mya.
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So,

We'll just use A to represent both Jan and Mya's miles, since they ran the same number.

We have the equations:
1. Jan (J) = Mya (M)
2. Sara (S) = M - 8
3. 2A + S = 64
J = M
S = M - 8

We'll just use A to represent both J and M.

S = M - 8

We'll use Elimination by Substitution.
2A + A - 8 = 64

Collect Like Terms
3A - 8 = 64

Add 8 to both sides
3A = 72

Divide both sides by 3
A = 24

Since
A = J
and
A = M
and
J = M
then
J = 24
M = 24

Substitute
S = 24 - 8
S = 16

Check
24 + 24 + 16 = 64
64 = 64 This checks.

So,
J = 24
M = 24
S = 16
4 0
3 years ago
Read 2 more answers
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