MATHEMATICS COLLEGE

The scores on a test are normally distributed. The mean of the test is 750 and the standard deviation is 70. By using the Empirical rule, what scores fall 3 standard deviations from the mean?

a.610 and 820

b.610 and 960

c.680 and 820

d.540 and 960

Answers

Answer 1
Answer: I think it will be B 610 and 820 but im not sure.
Hope This Helps!
~Cupcake

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Answers

Answer:

18

Step-by-step explanation:

2x-10=x+8

2x-x=8+10

x=18
Answer is 18 hoped this help

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function for . Determine the following (round all of your answers to 3 decimal places): (a) Enter your answer in accordance to the item a) of the question statement .972 (b) Enter your answer in accordance to the item b) of the question statement .0123 (c) Enter your answer in accordance to the item c) of the question statement .028 (d) Enter your answer in accordance to the item d) of the question statement .972 (e) Determine such that . Enter your answer in accordance to the item e) of the question statement

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

a   P(X <  5) =  0.960

b P(X >  8) = 0.016

c   P(6 < x < 10) =  0.018

d  P(X < 6 or  X > 10 ) =  0.982

e  X =  2

Step-by-step explanation:

From the question we are told that

   The probability density function is  f(x) =  (2)/(x^3) for  x > 1

Considering question a

  P(x < 5) = \int\limits^5_1 {(2)/(x^3) } \, dx

=> P(X <  5) =  [-(1)/(x^2) ]|  \left \ 5} \atop {1}} \right.

=>P(X <  5) = - (1)/(25)  +   (1)/(1^2)

=>  P(X <  5) =  0.960

Considering question b

P(x > 8) =1  - \int\limits^6_1 {(2)/(x^3) } \, dx

=> P(X > 8) =1-  [-(1)/(x^2) ]|  \left \ 8} \atop {1}} \right.

=>P(X >  8) = 1 - [- (1)/(64)  +   (1)/(1^2)]

=>P(X >  8) = 0.016

Considering question c

  P(6 < x < 10) = \int\limits^(10)_(6) {(2)/(x^3) } \, dx

=> P(6 < x < 10) =  [-(1)/(x^2) ]|  \left \ 10} \atop {6}} \right.  

=>P(6 < x < 10) =  [- (1)/(100)  +   (1)/(36)]

=>P(6 < x < 10) =  0.018

Considering question d

 P(X < 6 or  X > 10 ) = 1 - P(6 < x < 10) = 1 - \int\limits^(10)_(6) {(2)/(x^3) } \, dx

=> P(X < 6 or  X > 10 ) =1-  [-(1)/(x^2) ]|  \left \ 10} \atop {6}} \right.

=> P(X < 6 or  X > 10 ) =1- [- (1)/(100)  +   (1)/(36)] [/tex]

=> P(X < 6 or  X > 10 ) =  0.982

Considering question e

   P(X  <  x ) =  \int\limits^x_1 {(2)/(x^3) } \, dx  =  0.75

   P(X  <  x ) =  [- (1)/(x^2) ]| \left \ x } \atop {1}} \right.  =  0.75

   P(X  <  x ) =  - (1)/(x^2) - [- (1)/(1^2) ]= 0.75

   P(X  <  x ) =  - (1)/(x^2) + 1 = 0.75

     - (1)/(x^2)  = -0.25

     X =  2

     

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Answers

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Answers

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Answers

Answer:

172.12 sq in

Step-by-step explanation:

The figure is comprised of a semicircle, and 2 rectangles

Area of semicircle = 1/2(3.14)4^2 = 25.12

Area of rectangle with length of 9 and width of 7 = 7(9) = 63

Area of rectangle with length of 14 and width of 6 = 6(14) = 84

Total area = 25.12 + 63 + 84 = 172.12 sq in

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Answers

Answer:

C

Step-by-step explanation:

HJN IS SIMILAR TO PQN
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