A teacher asked all the students in the class which days of the week they get up after 8 a.m.

Which of the following is the best way to display the frequency for each day of the week?

In a negatively-correlated variables, a change in one variable result in opposite change in the other variable. In other words, an increase in on variable results in a decrease in the other.

Now let’s analyze the given scenarios:

As elevation decreases, temperature remains the same this is NO-CORRELATION

As elevation decreases, temperature decreases is a POSITIVE CORRELATION

As elevation increases, temperature decreases is a NEGATIVE CORRELATION

As elevation increases, temperature increases is a POSITIVE CORRELATION

Standard interior doors typically range from0.7 to 0.9 metersin width and 2.0 to 2.1 meters in height. Thus, the average width of the doorway is about 1.0 m.

To find a solution to this, we need to find the change in spending between the year 2013 and 2015 for each country. Change is difference in amount spent in 2015 and 2013 for each country:

Sudan=$(24-17)= $7

Eritrea=$(42-56)= - $14

Nigeria = $(21-16)= $5

Tanzania =$(81-68) = $13

Rwanda = $(24-27) =-$3

Kenya = $(88-81) = $8

A Minus sign means a decrease in spending. From the above calculations, Eritrea had the GREATEST change in dollars spent while Rwanda had the LEAST change in dollars spent.

Step 1: Find a common denominator

The least common denominator (LCD) of 6 and 3 is 6.

Convert 4/3​ to have denominator 6:

4/3=8/6

Step 2: Add the fractions

5/6+8/6=13/6​

Step 3: Convert to a mixed number

13/6 = 2 1/6

The best estimate is about 2 ½ cups.

The dimension of the field is shown below.

Area of rectangle = length * width

Area of rectangle = 100 m *75 m = 7500 m2

Area of the rectangle in hectares =

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/rangedata. While a bar graph does just fine even with discrete data.

Let the amount the friend pays be x. Then, 2 times what the friend pays is 2x. Less $2,000 what the friend pays is 2x-2000.

Now we know that

2x-2000=48000

Add 2000 to both sides of the equation

2x-2000+2000=48000+2000

2x+2000-2000=50000

2x=50000

Divide both sides by 2

X=25000

Thus, the friend pays $25,000 for their car.

In this case, we are measuring average weight of the babies with time. The weight of the babies will vary with time, meaning when we vary time, the average weight of the babies will be different. Here, time is independent while average weight is dependent variable.

Hire purchase cost=down payment + total monthly installment

Total monthly payment=hire purchase cost-down payment= $(1570-400)=$ 1170

Number of months=total monthly payment/monthly payment=$1170/$100=11.7 months

The consumer will clear the balance in 12 months.

Number of boxes in the room is equal to the volume of the room divided by the volume of one box

Volume of the room=9 ft*9 ft*9 ft= 729 ft3

Volume of one box = 2 ft* 2 ft * 2 ft= 8 ft3

Number of boxes= 729 ft3/ 8 ft3= 91.125 boxes

Thus, the room can store 91.125 boxes, which is about 91 boxes.

However, the room’s dimensions are 9 ft x 9 ft x 9 ft, and the boxes are 2 ft x 2 ft x 2 ft. To determine how many boxes fit along each dimension:

• In the 9 ft length, you can fit\(\frac{9}{2} = 4.5\) boxes, which means only 4 whole boxes fit along this dimension.
• In the 9 ft width, you can fit 4 whole boxes.
• In the 9 ft height, you can fit 4 whole boxes.

Therefore, the total number of boxes that fit is:

\(4×4×4=64\ boxes\)