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Virty [35]
3 years ago
12

Suppose you wish to test if a number cube (die) is loaded or not. If the die is not loaded, the theoretical probabilities for ea

ch roll should be: 1 2 3 4 5 6 16 2/3 % 16 2/3 % 16 2/3 % 16 2/3 % 16 2/3 % 16 2/3 % You roll the die 84 times and come up with the
Mathematics
1 answer:
Sonja [21]3 years ago
3 0

Answer:

Goodness of fit

Step-by-step explanation:

Given

The theoretical probabilities

<em>See comment for complete question</em>

<em></em>

Required

The type of test to be use

From the question, we understand that you are to test if the die is loaded or not using the given theoretical probabilities.

This test can be carried out using goodness of fit test because the goodness of fit is basically used to check the possibility of getting the outcome variable from a distribution. In this case, the outcome of the variables are the given theoretical probabilities.

In a nutshell, the goodness fit of test determines if the given data (in this case, the theoretical probabilities) is a reflection of what to expect in the original population.

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Choose all of the equivalent ratios for 64:80<br> 8:12<br> 32:40<br> 4:5<br> 16:24<br> 120:160
Alinara [238K]

Answer:

32/40, 4/5

Step-by-step explanation:

All these ratios are follow the ratio 4:5 meaning they are equivalent to "64:80".

8:12=0.666667

16"24=0.666667

120:160=0.75

5 0
3 years ago
Read 2 more answers
Write the slope-intercept form of the equation of each line.​
yanalaym [24]

Answer:

y=mx+b

y=x+3

Step-by-step explanation:

m: 1 (rise over run by 1)

b: 3

7 0
2 years ago
Alguien me pude decir una adivinanza o historia q la solución sea una ecuación de segundo grado completa????
Hitman42 [59]

Answer:

Una ecuación de segundo grado completa es algo como:

y = a*x^2 + b*x + c

Un problema tipico que usa ecuaciones de este tipo es algo que nos haga multiplicar dos términos como h*(x + k) y n*(x + j)

Donde h, k, n, y j son números reales.

Entonces planteemos el siguiente caso.

"Juan tiene una tabla rectangular, de tal forma que su largo es (x + 3cm), y el ancho es igual a dos veces su largo, si sabemos que el área de esta tabla es 95 cm^2 ¿cuál es el valor de x?"

Acá hay que recordar que para un rectángulo de largo L y ancho W, el área es:

A = L*W

En este caso tendremos:

L = (x + 3cm)

Y el ancho es dos veces eso, entonces:

W = 2*(x + 3cm)

Entonces el area de esta tabla se escribirá como:

A = (x + 3cm)*2*(x + 3cm)

Ahora solo tenemos que expandir eso:

A = 2*x^2 + 12cm*x + 18cm^2

Ahora sabemos que el area de esta tabla es 95cm^2

Entonces:

95cm^2 =  2*x^2 + 12cm*x + 18cm^2

0 = 2*x^2 + 12cm*x + (18cm^2 - 95cm^2)

0 = 2*x^2 + 12cm*x - 77 cm^2

Esta es una ecuación de segundo grado completa, la cual deberiamos resolver para encontrar el valor de x como se nos pide en el problema.

4 0
2 years ago
What is a net in math?
WITCHER [35]

Answer:

it is a two-dimensional shape that can be modified to form a three-dimensional shape or a solid

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Rework problem 35 from the Chapter 2 review exercises in your text, involving auditioning for a play. For this problem, assume 9
sweet-ann [11.9K]

Answer: 1) 6300 ways

2) 2520 ways

3) 0.067

Step-by-step explanation:

For this problem, assume 9 males audition, one of them being Winston, 5 females audition, one of them being Julia, and 6 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available.

How many different ways can these roles be filled from these auditioners?

Available: 9M and  5F and 6C

Cast: 3M and 1F and  2C

As it is not ordered: C₉,₃ * C₅,₁ * C₆,₂

C₉,₃ = 9!/3!.6! = 84

C₅,₁ = 5!/1!.4! = 5

C₆,₂ = 6!/2!.4! = 15

C₉,₃ * C₅,₁ * C₆,₂ = 84.5.15 = 6300

How many different ways can these roles be filled if exactly one of Winston and Julia gets a part?

2 options Winston gets or Julia gets it:

1) Winston gets it but Julia no:

8 male for 2 spots

4 females for 1 spot

6 children for 2 spots

C₈,₂ * C₄,₁ * C₆,₂

C₈,₂ = = 8!/2!.6! = 28

C₄,₁ = 4!/1!.3! = 4

C₆,₂ = 6!/2!.4! = 15

C₈,₂ * C₄,₁ * C₆,₂ = 28.4.15 = 1680

2) Julia gets it but Winston does not

8 male for 3 spots

1 female for 1 spot

6 children for 2 spots

C₈,₃ * C₆,₂

C₈,₃ = 8!/3!.5! = 56

C₆,₂ = 6!/2!.4! = 15

C₉,₃ * C₆,₂ = 56.15 = 840

1) or 2) = 1) + 2) = 1680 + 840 = 2520

What is the probability (if the roles are filled at random) of both Winston and Julia getting a part?

8 male for 2 spots

1 female for 1 spot

6 children for 2 spots

C₈,₂ * C₆,₂

C₈,₂ = = 8!/2!.6! = 28

C₆,₂ = 6!/2!.4! = 15

C₈,₂ * C₆,₂ = 28.15 = 420

p = 420/6300 = 0.067

6 0
3 years ago
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