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ASHA 777 [7]
3 years ago
5

Which is greater 0.7repeating or 3/4

Mathematics
2 answers:
Cerrena [4.2K]3 years ago
6 0
0.7 cause 3/4 is equivalent to 0.75 which is less
Kobotan [32]3 years ago
4 0

Answer:

0.7 repeating I believe

Step-by-step explanation:

i'm not sure how i got it, but yea

You might be interested in
I dont know how to do this
inn [45]

The dimensions of the rectangle are: length= 18 in and width =1 in.

<h3>Quadrilaterals</h3>

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle can be found for the formula : l*w, where b = length and w =width. The question gives that the area is 18 in².

For this question, the length exceeds its width by 17 inches - l=w+17. Thus,  from the value of area given, you can find the values of the length and width of the rectangle.

A=l*w

18=(w+17)*w

18=w²+17w

w²+17w-18=0

Next step will be solve the previous equation ( W²+17W-18=0)

w_{1,\:2}=\frac{-17\pm \sqrt{17^2-4\cdot \:1\cdot \left(-18\right)}}{2\cdot \:1}\\ \\ w_{1,\:2}=\frac{-17\pm \:19}{2\cdot \:1}

Therefore,

w_1=\frac{-17+19}{2\cdot \:1}=1\\ \\ \\ w_2=\frac{-17-19}{2\cdot \:1}=-18

For dimensions, only positive numbers must be used. Then, the width is equal to 1 inch.

As, the area (l*w) is 18 in², you have.

18=l*w

18=l*1

l=18 in

Read more about the area of rectangle here:

brainly.com/question/25292087

#SPJ1

3 0
2 years ago
Is 300 greater than 8/9?
koban [17]

Answer:

yes.

Step-by-step explanation:

300>8/9

300>0.888888889

8 0
3 years ago
Read 2 more answers
The ternary search algorithm locates an element in a list of increasing integers by successively splitting the list into three s
Zigmanuir [339]

Answer:

Algorithm

The steps involved in this algorithm are:

(The list must be in sorted order)

• Step 1: Divide the search space (initially, the list) in three parts (with two mid-points: mid1 and mid2)

• Step 2: The target element is compared with the edge elements that is elements at location mid1, mid2 and the end of the search space. If element matches, go to step 3 else predict in which section the target element lies. The search space is reduced to 1/3rd. If the element is not in the list, go to step 4 or to step 1.

• Step 3: Element found. Return index and exit.

• Step 4: Element not found. Exit.

Complexity

• Worst case time complexity: O(log N)

• Average case time complexity: O(log N)

• Best case time complexity: O(1)

• Space complexity: O(1)

Algorithm

The steps involved in this algorithm are:

(The list must be in sorted order)

• Step 1: Divide the search space (initially, the list) in three parts (with two mid-points: mid1 and mid2)

• Step 2: The target element is compared with the edge elements that is elements at location mid1, mid2 and the end of the search space. If element matches, go to step 3 else predict in which section the target element lies. The search space is reduced to 1/3rd. If the element is not in the list, go to step 4 or to step 1.

• Step 3: Element found. Return index and exit.

• Step 4: Element not found. Exit.

Complexity

• Worst case time complexity: O(log N)

• Average case time complexity: O(log N)

• Best case time complexity: O(1)

• Space complexity: O(1)

#include <stdio.h>

int ternarySearch(int array[], int left, int right, int x)

{

  if (right >= left) {

    int intvl = (right - left) / 3;

    int leftmid = left + intvl;

    int rightmid = leftmid + intvl;

    if (array[leftmid] == x)

       return leftmid;

    if (array[rightmid] == x)

       return rightmid;

    if (x < array[leftmid]) {

      return ternarySearch(array, left, leftmid, x);

    }

    else if (x > array[leftmid] && x < array[rightmid]) {

      return ternarySearch(array, leftmid, rightmid, x);

    }

    else {

      return ternarySearch(array, rightmid, right, x);

    }

  }

  return -1;

}

int main(void)

{

  int array[] = {1, 2, 3, 5};

  int size = sizeof(array)/ sizeof(array[0]);

  int find = 3;

  printf("Position of %d is %d\n", find, ternarySearch(array, 0, size-1, find));

  return 0;

}

Step-by-step explanation:

6 0
3 years ago
Let ​ x2+11x=3 ​.
IgorC [24]
X= 3/13. First you need to combine like terms, which would make 13x, and your equation would be 13x=3. then divide both sides by 13 and your X equals 3/13.
4 0
3 years ago
A polygraph (lie detector) is an instrument used to determine if the individual is telling the truth. These tests are considered
zhannawk [14.2K]

Answer:

Step-by-step explanation:

a) The probability of a Type I error in a lie detection test would be the probability that the lie detection machine incorrectly detected lie for the truth tellers. This is already given in the problem as 0.07.

Therefore,

P(Type-I) = 0.07

Therefore 0.07 is the required probability here.

b) The probability of a Type II error in a lie detection test would be the probability that the lie detection machine incorrectly detected truth for the the people who are actually liars. This is thus 1 - reliability.

P(Type-II) = 1 - Reliability = 1- 0.86 = 0.14

Therefore 0.14 is the required probability here.

7 0
3 years ago
Read 2 more answers
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