A bucket can hold 2500 mL. How many L can the bucket hold?

: In order to solve for the value of x, we use the order of operations as follows:

The first step is to divide both sides by 3

Add 2 to both sides to eliminate it from the left-hand side of the equation

x-2+2=4+2

x = 6

Thus, the value of x is 6.

We need to find how many mL are in 5.5 teaspoons. We are told that 1 teaspoon is about 4.93 mL, which we interpret as follows:

Or

Since we are looking for mL, we use the second option.

Thus, 5.5 teaspoons can hold approximately 27 mL.

To solve the equation this equation, we use a calculator and find the product as:

3.45 * 0.53= 1.8285

Explanation: We know that 1 L=1000 mL and to change between L and mL we use the following two options

We convert the given word problem into a mathematical expression.

Let the number be x.

Thrice a number = 3x

Twelve less than thrice a number = 3x-12

Thus, the required mathematical expression is 3x-12

Correlation of two variables falls into:

Positive correlation: an increase in one variable causes another variable to increase

Negative correlation: an increase in one variable causes another one to decrease

No correlation: a change in one variable does not cause any response in another variable.

From the given choices

Option a is no correlation

Option b is a negative correlation

Option c is a negative correlation

Option d is a positive correlation

Thus, the correct statement is a.

Before we compare the values of the options, we convert the decimals to fractions.

5.4 becomes 54/10

6.25 becomes 625/100

Thus, the set of fractions is 11/2, 54/10, 13/3, 625/100. The least common denominator for 2, 10, 3, and 100 is 300, which we use to compare each options given.

11/2*300=1650

54/10*300=1620

13/3*300=1300

625/100*300=1875

From the above, 1875 is the greatest value which corresponds to 6.25. Therefore, 6.25 is the greatest value of all the choices given.

From the given estimations, we can find the exact sum after adding the given numbers and see the trend the student utilizes to find their estimates. The actual sum is as follows

26+180=206

115+391=506

497+2024=2521

From the above evaluation, the students round the exact sum to the nearest tens. So, in the problem we need to approximate 545+7501.

545+7501=8046

The exact sum is 8046 and the approximate value after rounding to the nearest tens is 8050.

Here we need to find the actual values of the additions and see the trend the student will use to estimate the given problem. The exact sum is as follows

35+192=227

258+350=608

947+1032=1979

From the above calculations, it is evident that the student rounds up or down the ones place values. So, in the problem we need to approximate 694+7618.

694+7618=8312

We can approximate 8312 based on the provided choices is 8310.

we are asked to find the largest value from the options given. To solve this problem, we need to convert all the given values into fraction. Then find the least common denominator for all fractions, which we will use to multiply each fraction and make a comparison.

43÷55=43/55

5/7

0.729=729/1000

73%=73/100

Now, we need to find the least common denominator of 55, 7, 1000, and 100, which is 77000. We multiply each fraction by 77000 as follows:

43/55*77000=60200

5/7*77000=55000

729/1000*77000=56133

73/100*77000=56210

From the calculated values, it can be noted that 43/55 is the greatest value among the given options.