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oksian1 [2.3K]
3 years ago
10

Suppose in a large class, students' grade in a test followed the normal distribution. Approximatley what percent of grades is be

tween two standeard deviation below the mean and two standar deviation above the mean (that is in between Xbar-2S and Xbar+2S grades)? note, Xbar is the mean and S is standard deviation.
Mathematics
1 answer:
lilavasa [31]3 years ago
5 0

Answer:

95%

Step-by-step explanation:

Given that in a large class, students' grade in a test followed the normal distribution.

The normal distribution is symmetric about the mean, bell shaped and having area of more than 99% between 3 std deviations from the mean on either side.

We are to find the percent of grades is between two standeard deviation below the mean and two standar deviation above the mean (that is in between Xbar-2S and Xbar+2S grades)

where x bar is the mean and s = std devition

This is equivalent to 100*Prob that X lies between these two values

As per normal distribution curve 68, 95, 99 rule we get

approximately 95% lie between these two values.

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In a cylindrical wooden block with a radius of 7 cm and a height of 14 cm, hemispherical blocks with
Sunny_sXe [5.5K]

9514 1404 393

Answer:

  1372π/3 cm^3 ≈ 1436.8 cm^3

Step-by-step explanation:

The diameter and height of the cylinder are both 14 cm. The resulting solid will be a 14 cm sphere. Its volume is ...

  V = 4/3πr^3 = 4/3π(7 cm)^3 = 1372π/3 cm^3 ≈ 1436.8 cm^3

4 0
3 years ago
PLEASE HELP <br> Show work :)
S_A_V [24]

Step-by-step explanation:

a a a a a a a abedRounding of a number means replacing it with another number that is nearly equal to it but easy to represent or write. For example 754 rounded to nearest thousand is 1000. If the rounding number is 4 or less it is round down. If the rounding number is 5 or more it is rounded up

4 0
3 years ago
100 is 1/10 of what?
lawyer [7]

\boxed{ \ 100 \ is \ \frac{1}{10} \ of \ 1,000 \ }  

<h3>Further explanation  </h3>

The question can be rewritten to \boxed{ \ 100 \ is \ \frac{1}{10} \ of \ M \ }  

This case about equations with one variable. We have to solve this calculation to obtain the value of M. Our task is to isolate the variable M alone at the end of the process on one side of the equation until the variable will be equal to a value on the opposing side.  

Let's arrange it into an equation, i.e., \boxed{ \ 100 \ = \ \frac{1}{10} \times \ M \ }  

Turn this equation over so that the position of variable M is on the left side.  

\boxed{ \ M \times \frac{1}{10} = 100 \ }  

Both sides are multiplied by 10 to isolate M on the left side. The fraction are eliminated.  

\boxed{ \ M \times \frac{1}{10} \times 10 = 100 \times 10 \ }  

M = 1,000  

Therefore, we have calculated the number that was asked.  

Hence \boxed{ \ 100 \ is \ \frac{1}{10} \ of \ 1,000 \ }  

<u>Note:  </u>

The significant thing to carry out is how to manipulate both sides of the equation with the algebraic properties of equality such as:  

  • Adding  
  • Subtracting  
  • Multiplying, and/or  
  • Dividing both sides of the equation with a similar number

And the commutative property of multiplication, i.e., \boxed{ \ a \times b = b \times a \ }

<h3>Learn more  </h3>
  1. The similar case brainly.com/question/106300
  2. The similar case brainly.com/question/96882  
  3. Calculating mass based on density and volume brainly.com/question/4053884

Keywords: 100 is 1/10 of, solve, 1,000, variable, 2/7m - 1/7 = 3/14, algebraic properties of equality, one, linear equation, isolated, manipulate, operations, add, subtract, multiply, divide, fraction, equate, denominator, numerator, both sides, equal, the opposite, both sides are multiplied by, turn, calculation  

7 0
3 years ago
The PTA sells school supply packs each year. If 60% of the 400 students purchase from the PTA, then how many school supply packs
Bess [88]

:} Im sorry i dont know

...........................................................

8 0
2 years ago
You would 'b' be in '-20= -8(4)+b'
Volgvan
It’s -11 becuase i just took the test and got it
7 0
3 years ago
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