Write an equation perpendicular to the x axis and contains the points (9,-2)

The state of mishgan recored an all time high snowball of 355.9 inches during the winter of 1978-79 its lowest snowfall of 81.3

Rama09 [41]

Answer:

<u>274.6 inches</u> more snowfall Michigan received during the winter of 1978-79 than in 1930-31.

Step-by-step explanation:

Given:

The state of Michigan recorded an all time high snowfall of 355.9 inches during the winter of 1978-79.

Its lowest snowfall of 81.3 inches was recorded during the winter of 1930-31.

Now, to find how much more snowfall did Michigan received during the winter of 1978-79 than in 1930-31.

The snowfall in 1978-79 = 355.9 inches.

And, the snowfall in 1930-31 = 81.3 inches.

So, to get the how much more snowfall in 1978-79 than 1930-31 we subtract:

Snowfall in 1978-79 - Snowfall in 1930-31.

$355.9\ inches-81.3\ inches = 274.6\ inches.$

Therefore, 274.6 inches more snowfall Michigan received during the winter of 1978-79 than in 1930-31.

8 0

3 years ago

Evaluate the numerical expression of 27 + 5 × 6

sesenic [268]

You have to use PEMDAS, by multiplying before you add. So when you multiply 5*6, you get 30 and add the remaining 27 to get 57.

4 0

3 years ago

If 2X +3 equals 11 what is 4X hint solve for x first

Aleksandr [31]

Answer:

16

Step-by-step explanation:

2x+3 = 11

2x= 8

x = 4

4(x) = 4(4) = 16

7 0

3 years ago

Using the Breadth-First Search Algorithm, determine the minimum number of edges that it would require to reach

jekas [21]

Answer:

The algorithm is given below.

#include <iostream>

#include <vector>

#include <utility>

#include <algorithm>

using namespace std;

const int MAX = 1e4 + 5;

int id[MAX], nodes, edges;

pair <long long, pair<int, int> > p[MAX];

void initialize()

{

for(int i = 0;i < MAX;++i)

id[i] = i;

}

int root(int x)

{

while(id[x] != x)

{

id[x] = id[id[x]];

x = id[x];

}

return x;

}

void union1(int x, int y)

{

int p = root(x);

int q = root(y);

id[p] = id[q];

}

long long kruskal(pair<long long, pair<int, int> > p[])

{

int x, y;

long long cost, minimumCost = 0;

for(int i = 0;i < edges;++i)

{

// Selecting edges one by one in increasing order from the beginning

x = p[i].second.first;

y = p[i].second.second;

cost = p[i].first;

// Check if the selected edge is creating a cycle or not

if(root(x) != root(y))

{

minimumCost += cost;

union1(x, y);

}

return minimumCost;

}

int main()

{

int x, y;

long long weight, cost, minimumCost;

initialize();

cin >> nodes >> edges;

for(int i = 0;i < edges;++i)

{

cin >> x >> y >> weight;

p[i] = make_pair(weight, make_pair(x, y));

}

// Sort the edges in the ascending order

sort(p, p + edges);

minimumCost = kruskal(p);

cout << minimumCost << endl;

return 0;

}

8 0

3 years ago

If an event has a 55% chance of happening in one trial, how do I determine the chances of it happening more than once in 4 trial

aev [14]

The chances of it happening more than once in 4 trials is 13%

<h3>How to determine the number</h3>

From the information given, we have can deduce that;

Probability of 1 trial = 55%

= 55/ 100

Find the ratio

= 0. 55

We are to find the probability of it happening more than once in 4 different trials

If the probability of it happening in one trial is 555 which equals 0. 55

Then the probability of it happening in 1 in 4 trials is given as;

P(1/4 trials) = 1/ 4 × 55%

P(1/4 trials) = 1/ 4 × 0. 55

Put in decimal form

P(1/4 trials) = 0. 25 × 0. 55

P(1/4 trials) = 0. 138

But we have to know the percentage

= 0. 138 × 100

Multiply the values, we have

= 13. 8 %

Thus, the chances of it happening more than once in 4 trials is 13%

Learn more about probability here:

brainly.com/question/24756209

#SPJ1

6 0

2 years ago