An athlete can run 6 miles in 51 minutes. At this rate, how many miles could the athlete run in 90 minutes?

To answer this question, we need to observe the trend of the number of days with dollars. On day 2, the dollars were 8, after 5 days, the dollars decreased to 3, after 13 days, the dollars increased to 10. After 14 days, the dollars declined to 6 and again rose to 16 dollars after 19 days.

From this observation, we cannot be able to establishany correlation between days and dollars.

Here we are required to find the area of the square of sides 3.1 m. The square is a four-sided figure with each side equal and opposite sides making 90 degrees.

Area of the square =side*side

Side=3.1 m

Area of the square =3.1 m *3.1 m=9.61 m2

Note: 3.1 + 3.1 = 6.2 which is a wrong answer.

Here we collect like terms together and solve for the unknown value of x.

7x-6=3x-26

Add 6 to both sides of the equation

7x-6+6=3x-26+6

7x=3x-20

Subtract 3x from both sides of the equation

7x-3x=-20

4x=-20

Divide both sides by 4

4x/-4=-20/4

x = -5

The value of x = -5

In the simple interest, we utilize the following formula to find the simple interest after a period of time in years.

I=P*r*t

I is the interest

P=Principal or initial deposit

r=rate

t=time in years

From the given problem, P=$600, r=6%=6/100=0.06, t=5 years. Then

I=$600*0.06*5=$180

After 5 years, Pat will earn an interest of $180.

In this problem, we take the triangle as a right-angled triangle and label it as follows:

From the Pythagoras theorem A2 + B2= C2, we can look for a combination of A and B that when the squares of A and B are summed give a square of 13.  Mathematically,

But C=13 inches

If we take A=5 inches and B=12 inches, then

Next, we take A=2.5 inches, B=6 inches

Next, we take A=2.5 inches, B=4 inches

6.25+16=169

22.25≠169

Taking A=5 inches and B=8 inches

From the above computation, the combination of A=5 inches and B=12 inches give a triangle with a hypothenuse of 13 inches.

We solve the problem by first finding the amount of gallons needed for the whole trip then find the cost of the gallon.

The car uses 1 gallon to cover 25 miles, which is same as:

Now, in 3000 miles, the car will use how many gallons of gasoline?

A trip of 3000 miles will need 120 gallons. And to find the cost of 120 gallons, we proceed as follows:

The family will spend $199.20 to drive for 3000 miles.

we are asked to change 400 mg to g. We use the conversion 1g=1000mg, which we can write as:

So, 400 mg to g becomes

From the given problem,

49.5 pounds of fertilizer is needed to farm 1 acre of land. This can be interpreted as:

Or

Now we are needed to find the acres of land that will be farmed using 2000 pounds of fertilizers. To solve this, we use the second option as follows:

2000 pounds of fertilizers can farm approximately 40 acres of land.

We extrapolate the values of a line between 10 and 15 on y-axis to meet the curve and drop another line from the intersection of the curve to meet the x-axis and read off the age of the car.

Now, let’s extrapolate in the following diagram

The intersection of the line and the x-axis is between 4 and 5 years. This means that the car is about 4.5 years old. From the options given, the age of the car is more than 3 years old.

to do a comparison of the given data set, we convert the fraction numbers into the approximate decimals.

4/3=1.333

2720=1.35

Now, we have 1.333, 1.35, 1.369, 1.357. These numbers are arranged from the smallest to the largest number, we look at the number in the second decimal place value

1.333 the number in second decimal place value is 3

1.35 the number in third decimal place value is 5

1.369 the number in third decimal place value is 6

1.357 the number in third decimal place value is 5

However, 1.35 and 1.357 have 5 in the second-place vale. We use the ird place value to compare the two number:

1.35 is same as 1.350 and the number in the third-place value is 0

1.357 the number in the third-place value is 7

Now, 0 is less than 7, and thus 1.35 is less than 1.357

Overall, 3 is less than 5, and 5 less than 6 and the provided numbers can be arranged from the smallest to the largest as:

1.333, 1.35, 1.357, 1.369

Thus, the corresponding given numbers organized from the smallest to the largest become:

4/3, 27/20, 1.357, 1.369

The greatest value is 1.369