A student constructs a frequency distribution where the frequency of the first class is 8 and a corresponding histogram with the area of the bar in the first class is 80. The frequency for the second class is twice the frequency of the first class. In the histogram, the area of the bar corresponding to the second class is four times the area of the bar corresponding to the first class. What is the frequency of the second class and the area of the bar corresponding to the second class?

Question

A student constructs a frequency distribution where the frequency of the first class is 8 and a corresponding histogram with the area of the bar in the first class is 80. The frequency for the second class is twice the frequency of the first class. In the histogram, the area of the bar corresponding to the second class is four times the area of the bar corresponding to the first class. What is the frequency of the second class and the area of the bar corresponding to the second class?

Answer 1

frequency: 2*8=16

area: 4*80=320

Answer 2

Why did the student's frequency distribution and histogram go to therapy?

Because their relationship was completely imbalanced!

Now let's crunch some numbers. If the frequency of the first class is 8, then the frequency of the second class would be twice that, which is 16.

As for the areas of the bars in the histogram, if the area of the bar corresponding to the first class is 80, then the area of the bar corresponding to the second class would be four times that, which is 320.

So, the frequency of the second class is 16 and the area of the bar corresponding to the second class is 320.

Answer 3

To find the frequency of the second class, we know that it is twice the frequency of the first class. Therefore, we can set up an equation:

Let the frequency of the first class be x.
Frequency of the second class = 2x.

Now, let's find the area of the bar corresponding to the second class. We are given that the area of the bar for the first class is 80.

Let the area of the bar corresponding to the second class be y.
Area of the bar corresponding to the first class = 80.
Area of the bar corresponding to the second class = 4 times the area of the first class = 4 * 80 = 320.

So, the frequency of the second class is 2x, and the area of the bar corresponding to the second class is 320.

Answer 4

To solve this problem, we can use the information given to set up a system of equations. Let's denote the frequency of the first class as "f1" and the frequency of the second class as "f2." Similarly, let's denote the area of the bar corresponding to the first class as "a1" and the area of the bar corresponding to the second class as "a2."

According to the information given:
1. The frequency of the first class is 8: f1 = 8.
2. The frequency of the second class is twice the frequency of the first class: f2 = 2f1.
3. The area of the bar corresponding to the first class is 80: a1 = 80.
4. The area of the bar corresponding to the second class is four times the area of the bar corresponding to the first class: a2 = 4a1.

Now, let's solve for f2 and a2 using these equations:

1. We know that f1 = 8.

Substituting this into equation 2, we get:
f2 = 2(8)
f2 = 16.

So, the frequency of the second class is 16.

2. We know that a1 = 80.

Substituting this into equation 4, we get:
a2 = 4(80)
a2 = 320.

So, the area of the bar corresponding to the second class is 320.

Therefore, the frequency of the second class is 16, and the area of the bar corresponding to the second class is 320.