The scatter plot below shows the relationship between the number of hours a student plays golf and the student’s class attendance. Which of the following types of correlation is shown on the scatter plot?

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.

We need to find the amount Lana left after spending and investing another.

Lana spends=70% of $90=70/100 * 90 = $63

Amount left after spending=$(90-63) = $27

Lana is left with $27, which she will invest and earns a profit of 60%.

Profit earned=60% of $27 = 60/100 * 27 = $16.20

Therefore, Lana will have $27 + $16.20 = $43.20

We are required to find the doctor’s earning in 4 weeks after taking off taxes.

The first step is to find the total weekly deductions as follows:

Total weekly tax=federal income tax + state income tax + Social Security tax

Total weekly tax=$(85.00+40.00+81.00)

Total tax=$206.00

The next step is to find the weekly income after deducting total taxes

Weekly net income=gross income-total tax

Weekly net income=$(980.00-206.00)=$774.00

In one week, the net income of the doctor is $774.00 and after 4 weeks, the net income will be 4 times his weekly earning

In this scenario, we use the given scale in the charts to find the number of each coin and selected the right number of coins the student collected.

From the above evaluations, option two is the correct choice.

In this problem, we need to find the amount of gratuity the couple will leave. The gratuity is 16% of the total bill. Before solving the problem, the following are terms and their meaning in percent problems:

• Is means equals
• Of means multiply
• What means unknown (variable)

If we let x be the amount of gratuity, then translating the given problem into a mathematical equation becomes:

Now we evaluate the above equation noting that of means multiply.

So, the value of x=9.376 and to the nearest cent, x=9.38

There, a couple will leave a gratuity of $9.38.

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.

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.

The selling price of the sweater is marked price less the mark off value.

Mark off value=mark off rate*original price

Mark of value=30% of $95

Mark of value=0.30*$95=$28.50

Now, find the selling price of the sweater

Selling price= marked price-mark off value

Thus, the sweater will be sold for 66.50 after applying a mark off of 30%. The selling price is about $67.

We two conversions when asked to convert between kg and lb

And

We need kg equivalent of 220 lb, therefore we use option 1 and carry out the conversions as follows:

Thus, 100kg is equal to 220lb.

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.