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

Here, the athlete runs 6 miles in 51 minutes, which can be expressed as:

Now, in 90 minutes, the athlete will cover about

Therefore, the athlete runs about 10.6 miles in 90 minutes.

When dividing fractions, the division sign is changed into multiply sign and the first fraction is multiplied by the inverse of the second fraction.

Change division to multiply sign and find invers of 5/8. The inverse of a fraction is inverting the given fraction. The denominator becomes the numerator and the numerator becomes the denominator.

A whole pizza is equivalent to 1. Then

One friend eats half of the pizza=1/2 of 1=1/2*1=1/2

Remaining amount of pizza to be shared=1-1/2=2/2-1/2=1/2

Now, three friends share ½ equally and each friend will get,

Thus, the three friends each share 1/6.

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.

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.

The mean is the total divided by the number of elements in the data set. From the given data set:

Total=30+29+28+30+24+12+26+33+25+23=260

Number of items, N=10

Mean=total/N=260/10=26

The mean is 26.

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

We can interpret ‘cannot exceed” as less than ‘<’. Therefore, in our inequality, the symbol < must be included. Now let’s convert the word problem into a mathematical inequality.

Money spent on supplies=s

Money spent on textbooks=t

Total money spent=money spent on supplies + money spent on textbooks

Total money spent = s+t

But the money spent cannot exceed $12,000. Then,

s+t <$12,000.
However the money spent can still be equal to $12000, as it has not exceeded it.

Therefore, therequired inequality is s + t <= $12,000.

The best way to display the frequency of each day of the week when students get up after 8 a.m. is by using a bar graph. Bar graphs are well-suited for representing categorical data, where each day of the week is a separate category, and the height of each bar corresponds to the count or frequency of students waking up late on that specific day.

Note: Histograms, on the other hand, are more appropriate for visualizing continuous or numerical data and are not ideal for categorical data like days of the week.Histograms are useful for understanding the distribution of data, identifying patterns, and assessing the shape of the data distribution, such as whether it's normally distributed, skewed, or has multiple modes.

As you can see below, the Histogram is used to depict a pattern/continuous data. While a bar graph does just fine even with discrete data.

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.