The graph below represents the amount of rainfall in a particular state by month. Which of the following is the total rainfall for the first 5 months of the year?

In a bell-shaped curve, the data distribution is symmetric around a single peak. The centering of data around a single peak means the mean, mode and median of the test are all equal to each other.

To solve this problem, we need to have all options in fraction form.

74  95 = 

7/8 = 3/8

0.72 = 72/100

74% = 74/100

Now, find the least common denominator of 95, 8, and 100. The least common denominator is 3800, which we multiply by each fraction as follows:

74/95*3800=2960

7/8*3800=3325

72/100*3800=2736

74/100*3800=2812

We can see that the fraction 7/8 is the greatest among the given options.

Based rainfall falling for the first 5 months, a bar graph will give a great display for the farmer to decide when start planting. The horizontal scale will represent months of the year and vertical scale will present the amount of rainfall falling.

We use the order of operation to solve for the unknown value of x.

3(3x+3)=8x+5

Multiply 3 with each number in the brackets

(3*3x)+(3*3)=8x+5

9x+9=8x+5

Subtract 9 from both sides

9x+9-9=8x+5-9

9x=8x-4

Subtract 8x on both sides

9x-8x=8x-8x-4

x=-4

Thus, the unknown value of x is -4.

We need to form a mathematical expression from the given word problem.

Let the unknown number be x.

Thrice a number is three times = 3x

Twelve less than thrice a number = 3x-12

Thus, the mathematical expression from the word problem is 3x-12

We use given information to find how much ammonia is need to make the specified solution.

We are told, one gallon of cleaning solution requires 10oz of ammonia. Expressing this mathematically yields two options:

\(\frac{10\ oz\ of\ ammonia}{1\ gallon\ of\ solution}\)or\(\frac{1\ gallon\ of\ solution}{10\ oz\ of\ ammonia}\)

Now we find how much ammonia is needed using option two.

\(51\ gallon\ of\ solution\ *\frac{10\ oz\ of\ ammonia}{1\ gallon\ of\ solution}\ =\ 510\ oz\ of\ ammonia\)

From the above equation, gallon of solution will cancel, and oz of ammonia is left.

Therefore, the solution will require 510 oz of ammonia.

In order for the teacher to better visualize the test performance for his class, he needs to present the mark distribution in chart form. A better chart visualizer is the histogram, which will show the frequency of marks against range of test scores. A bar graph will not be used because the rectangles need to touch one another.

In a bar graph, the rectangles do not touch each other. Therefore, it will not be a good chart visualizer.

To solve the equation this equation, we compute the multiplication of the two numbers using the calculator

3.60 * 0.75 = 2.7

to find the average time the athlete runs in 1 mile; we find total distance and total time the athlete takes in running.

Total distance run by athlete=7+3=10 miles

Total time=45+15 =60 minutes

The athlete runs 10 miles in 60 minutes. For the athlete to cover 1 mile, he will run for a time less than 60 minutes. Thus, in 1 mile, the athlete takes:

So, in 6 minutes, the athlete covers 1 mile.

The initial step is to convert the decimal numbers into fractions.

5.2 becomes 52/10

5.15 becomes 515/100

Next, find the least common denominator of 3, 10, and 100, which is 300. We use this value and multiply each fraction.

7/3*300=700

52/10*300=1560

13/3*300=1300

515/100*300=1545

From the above, 1560 is the greatest value which corresponds to 52/10. Thus, 5.2 is the greatest value of all the choices given.