936 round to the nearest ten

Answers

Answer 1

Answer: It be rounded to 940

Answer 2

Answer: The answer would be 940. Since 6 is more than 5, you round it up.

Related Questions

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What does x equal? Plz hurry. First to answer correctly will be marked brainliest ASAP.
Devon is asked to draw a triangle with the following specifications:at least one angle measuring 60°when the angles are placed in order from least measure to greatest measure, the differences in measures are equalthe longest side of the triangle is 2 inchesWhich of the following statements about this triangle is true?A. Exactly one triangle exists with the given conditions, and it must be a right triangle.B. Exactly one triangle exists with the given conditions, and it must be an equilateral triangle.C. Exactly one triangle exists with the given conditions, and it must be a scalene triangle.D. More than one triangle exists with the given conditions, and there is not a specific type that all instances must be.

Find Four constructive odd integers whose sum is 224 what is the largest number

Answers

Final answer:

Given 4 consecutive odd numbers with a sum of 224, we can create an equation to solve for the first number (x). After calculating x = 53, we find the largest number to be 59.

Explanation:

This is a problem about consecutive odd integers, which are odd integers that follow each other successively. Let's denote the first odd integer as x, the second as x+2, the third as x+4, and the fourth as x+6. These are all odd because adding 2 to an odd number always results in the next odd number.

The question states that the sum of these four consecutive odd numbers is 224, so we can create the following equation: x + (x+2) + (x+4) + (x+6) = 224.

Simplify this equation to obtain 4x + 12 = 224. Then, solve for x by first subtracting 12 from both sides to get 4x = 212 and then dividing by 4 to get x = 53. Our highest odd number would be x+6 which equals 59.

Learn more about Odd Integers here:

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Final answer:

The four consecutive odd integers whose sum is 224 are 53, 55, 57, and 59. The largest of these integers is 59.

Explanation:

To find four constructive odd integers whose sum is 224, you first need to understand that consecutive odd integers follow a certain sequence, each increasing by 2 from the previous one. Let's label the first consecutive odd integer as 'n'. So, the four consecutive numbers will be: n, n + 2, n + 4, n + 6.

According to the problem, the sum of these four numbers equals 224:
n + (n+2) + (n+4) + (n+6) = 224.
Solve this equation to find the value of 'n'. You'll get n = 52. Hence, the numbers are 52, 54, 56, 58.

Since we are looking for odd numbers, let's start with 53 (the first odd number greater than 52) and proceed: 53, 55, 57, 59. And these four numbers do sum up to 224. Therefore, the largest number in this series is 59, our final answer.

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Which is greater negative 9 over 13 or 1and two thirds

Answers

1 and 2/3 this is because any positive is ALWAYS greater than a negative.

Hope this helped!!

one and two thirds because a positive number is always greater than a negative one

Which decimal is not equivalent 37.6 37.060 37.0600 37.06

Answers

37.6 because it doesn't matter how many zeros you put on after the umber it matter where the 6 is placed

37.6 because it doesnt matter how many zeros you put on after the umber it matter were the 6 is placed

Archie rolls two number cubes, each with sides numbered 1 through 6. He finds the sum of the numbers on the tops of the cubes. Whaf two sums have the probability?A. 3 and 4
B. 5 and 9
C. 5 and 8
D. 10 and 12

Answers

$\Omega=\{(x;y):x;y\in\{1;\ 2;\ 3;\ 4;\ 5;\ 6\}\}\n\n\overline{\overline{\Omega}}=6^2=36\n\nA=\{(1;2);\ (2;1);\ (1;3);\ (3;1);\ (2;\ 2)\};\ \overline{\overline{A}}=5;\ P(A)=(5)/(36)\n\nB=\{(1;4);\ (4;1);\ (2;3);\ (3;2);\ (3;6);\ (6;3);\ (4;5);\ (5;4)\};\n\overline{\overline{B}}=8;\ P(B)=(8)/(36)\n\nC=\{(1;4);\ (4;1);\ (2;3);\ (3;2);\ (2;6);\ (6;2);\ (3;5);\ (5;3);\ (4;4)\}\n\overline{\overline{C}}=9;\ P(C)=(9)/(36)$

$D=\{(4;6);\ (4;6);\ (5;5);\ (6;6)\};\ \overline{\overline{D}}=4;\ P(D)=(4)/(36)$

$A.\n3\to(1;2);\ (2;1)\to(2)/(36)\n4\to(1;3);\ (3;1);\ (2;2)\to(3)/(36)\n\nB.\n5\to(1;4);\ (4;1);\ (2;3);\ (3;2)\to(4)/(36)\n9\to(3;6);\ (6;3);\ (4;5);\ (5;4)\to(4)/(36)$

$C.\n5\to(1;4);\ (4;1);\ (2;3);\ (3;2)\to(4)/(36)\n8\to(2;6);\ (6;2);\ (3;5);\ (5;3);\ (4;4)\to(5)/(36)\n\nD.\n10\to(4;6);\ (6;4);\ (5;5)\to(3)/(36)\n12\to(6;6)\to(1)/(36)$

The same probability*

Write firstly what results it's able to get and their sum:
.....1....2.....3.....4....5.....6
1   2 3 4 5 6    7
2   3 4 5 6 7    8
3   4 5 6 7 8 9
4   5 6 7 8 9 10
5   6 7 8 9 10 11
6 7 8 9 10    11 12

As you can see there are four options to get "5" and also four options to get "9".
So the B is proper answer.
Sum 5 and sum 9 have the same probability.

Which decimal is equal to 7/10?

A. 0.71

B. 0.7

C. 0.07

D. 0.071