A farmer is planning for a growing season and want to know the trend in amount of rainfall in his area. He obtains rainfall falling in the first 5 months of the year as shown in the above table. Which of the following would best display the data?

Month	January	February	March	April	May
Rainfall (mm)	20	6	14	28	34

 

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.

The first five months of the year are January, February, March, April, and May. We use to graph to find the amount of rainfall in each month as follows:

In the month of January, rainfall was between o and 1, which is halfway. Therefore, rainfall in January is 0.5 inches

On February, rainfall was 1 inch and on March the rainfall was 2 inches.

On April, the rainfall amount was 4 inches while on month of May, the rainfall was about 3.5 inches. Therefore, total rainfall in the first 5 months was:

Total rainfall=0.5 inches+1 inches+ 2 inches+4 inches + 3.5 inches =11 inches

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 1 L =1000 mL to convert between the two units. The conversion options of interconversions are:

And

We want millimeters, use the second option to change L to mL as:

Therefore, 2.5 L is equivalent to 2500 mL.

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.

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.

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.

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+192227

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 required to find the percent equivalent of the given fraction.

We multiply 7/8 by 100 to convert it to percent. You convert fraction to percent.

Thus 7/8 is equal to 87.5%.

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.