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2 years ago

The range of scores in each interval of a grouped frequency distribution is called the

1 answer:

3 0

The answer is class intervals. A big set of data are grouped into different classes to get a hint of the distribution, and the range of such class of data is known as the Class Interval. In other words, these are range of scores in a group frequency distribution. Class intervals are commonly equal in width and are mutually exclusive. The middle of an interval is called a class mark and the ends of a class interval are called class limits. To calculate the class interval, divide the range by the number of classes.

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Find the polynomial equation of least degree with roots -1, 3, and (+/-)3i

jasenka [17]

Each of these roots can be expressed as a binomial:

(x+1)=0, which solves to -1
(x-3)=0, which solves to 3
(x-3i)=0 which solves to 3i
(x+3i)=0, which solves to -3i
There are four roots, so our final equation will have x^4 as the least degree

Multiply them together. I'll multiply the i binomials first:
(x-3i)(x+3i) = x²+3ix-3ix-9i²
x²-9i²
x²+9 [since i²=-1]

Now I'll multiply the first two binomials together:
(x+1)(x-3) = x²-3x+x-3
x²-2x-3
Lastly, we'll multiply the two derived terms together:

(x²+9)(x²-2x-3) [from the binomial, I'll distribute the first term, then the second term, and I'll stack them so we can simply add like terms together]

x^4 -2x³-3x²
 +9x²-18x-27
x^4-2x³+6x²-18x-27

7 0

2 years ago

In each represent the common form of each argument using letters to stand for component sentences, and fill in the blanks so tha

arlik [135]

Answer:

$a \to b$

$\neg b$

$\neg a$

If all prime numbers are odd, then 2 is odd.

2 is not odd.

Therefore, it is not the case that all prime numbers are odd.

Step-by-step explanation:

Considering part A

The first sentence is

If all computer programs contain errors, then this program contains an error.

The second sentence is

This program does not contain an error.

The third sentence is

Therefore, it is not the case that all computer programs contain errors.

Now we will use a and b as the letter to denote the component sentences and $\neg a$ and $\neg b$ when the component sentences is not the case

So a = If all computer programs contain errors

and

b = this program contains an error.

Generally then is represented as $\to$

Hence the first sentence is denoted as

$a \to b$

The second sentence is represented as

$\neg b$

The third sentence is represented as

$\neg a$

So part a can be represented as

$a \to b$

$\neg b$

$\neg a$

Considering part B

Here the objective is to fill in the blank spaces so that the logic of the sentence in part b is as that in part a

Now comparing the second statement a and b

"This program does not contain an error" = "2 is not odd" = $\neg b$

Hence "this program contains an error." = "2 is odd" = b

Now comparing the third statement a and b

Therefore, it is not the case that all computer programs contain errors.

Therefore, it is not the case that all prime numbers are odd

$\neg a$

Hence

If all computer programs contain errors, = if all prime numbers are odd.= a

3 0

2 years ago

In the diagram below DE is parallel to XY. What is the value of y?

jeka94

it would be 94 hope this helps

7 0

2 years ago

Read 2 more answers

One student ate 3/20 of all candies and another 1.2 lb. The second student ate 3/5 of the candies and the remaining 0.3 lb. What

VARVARA [1.3K]

Let us say there are x number of candies .

For first student,

He ate 3/20 of all candies, so 3/20 (x)

He also ate 1.2 lb of candies.

Total candies first student ate = (3/20) x +1.2

Second student ate 3/5 of all candies that is (3/5)x

He also ate 0.3 lb of candies.

Total candies second student ate = (3/5)x +0.3

Since total candies for both will be same, so equating the two expressions

$\frac{3}{20} (x) +1.2 = \frac{3}{5} x +0.3$

x=2

So there are 2 lb of candies in total.

First student ate = 0.3 +1.2 = 1.5lb of candies.

Second student ate = 1.5 lb of candies

8 0

2 years ago

Read 2 more answers

Kim drove from Mathtown at (-2, 5) to Geometryville at (3, -1) to Algebra Springs at (-6, -5), and then back to Mathtown. Find t

bogdanovich [222]

Answer:

Option (3)

Step-by-step explanation:

To solve this question we will use the formula,

Distance (d) between two points $(x_1,y_1)$ and $(x_2,y_2)$ ,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between Mathtown (-2, 5) to Geometryville (3, -1)

$d_1=\sqrt{(3+2)^2+(-1-5)^2}$

$=\sqrt{25+36}$

$=\sqrt{61}$

$=7.81$

Distance between Geometryville (3, -1) and Algebra Springs (-6, -5)

$d_2=\sqrt{(3+6)^2+(-1+5)^2}$

$=\sqrt{81+16}$

$=\sqrt{97}$

$=9.85$

Distance between Algebra Springs (-6, -5) and Mathtown (-2, 5)

$d_3=\sqrt{(6-2)^2+(5+5)^2}$

$=\sqrt{16+100}$

$=\sqrt{116}$

$=10.77$

Now total distance traveled by Kim = $d_1+d_2+d_3$

= 7.81 + 9.85 + 10.77

= 28.43

Therefore, Option (3) will be the answer.

7 0

2 years ago