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Lemur [1.5K]
3 years ago
6

You have a globe that has a radius of 11 inches. Find the volume of your globe

Mathematics
2 answers:
kirill [66]3 years ago
6 0

Answer:5,575.3in cubed

Step-by-step explanation:

nordsb [41]3 years ago
5 0

Answer:

V = 4/3

Step-by-step explanation:

Well first you have to pick a letter and i picked V. Then you have to calculate 11. Then you have an equation 11 +1. That equals 12. Then you find 2 numbers that equal 12. The 2 numbers are 4 and 3.

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"The graph of F(x), shown below, resembles the graph of G(x) = x2, but it has been changed somewhat. Which of the following coul
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G(x)=x²
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x(x-4)(x-1)

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3 years ago
COMO SE SACA UNA TABLA DE FRECUENCIA?????=? URGENTE ES PA HOY
xz_007 [3.2K]

Answer:

Para construir una tabla de frecuencias, procedemos de la siguiente manera:

Construye una tabla con tres columnas. La primera columna muestra lo que se organiza en orden ascendente (es decir, las marcas). ...

Repase la lista de marcas. ...

Cuente el número de marcas de conteo para cada marca y escríbalo en la tercera columna.

8 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

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  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
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