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stira [4]
3 years ago
7

Which expressions would complete this equation so that it has infinitely many solutions?

Mathematics
1 answer:
andrezito [222]3 years ago
8 0
The way I would do this is to distribute the original piece and get
8 + 16x -12, which is the same as 16x -4, so C is definitely one of the answers
Next I would distribute A, which isn't an answer because 8x +14 does not equal 16x + 4.  Finally I would distribute D, which comes out as 16x -4, so it is one of the answers(:
ANSWERS: C and D
I really hope this helped(:
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10
Alona [7]
The answer is 14.3%


because if u times it by 7 u get 100 because there is 1/7 percent chance it it will be sunday then any other day of the week
5 0
2 years ago
A box contains 35 gems, of which 10 are real diamonds and 25 are fake diamonds. Gems are randomly taken out of the box, one at a
Sonja [21]

Answer:

Step-by-step explanation:

The total number of ways of selecting 3 gems of any kind without replacement is 35 * 34 * 33 = 39270

Now selecting 2 gems from 25 is 25C2

25 C 2 = 25 * 24/2 = 300

Selecting one good gem is 10

Total selection = 300 * 10 = 3000

P(gems) = 3000 / 39270

P(gems) = 300/3927

P(gems) = 100/1309

6 0
2 years ago
-3/5 x = 18 what is x
Liono4ka [1.6K]
My guess would be x = -30

4 0
3 years ago
Read 2 more answers
Students are making ornaments and each takes 1 2 of a piece of Bristol board. They go to the cupboard and find 1 1 2 pieces of B
s2008m [1.1K]

Answer: 3

Step-by-step explanation:

Feom the question, we are informed that students are making ornaments and each takes 1/2 of a piece of Bristol board and that they go to the cupboard and find 1 1/2 pieces of Bristol board.

The number of ornaments that they can make will be calculated by dividing 1 1/2 by 1/2. This will be:

= 1 1/2 ÷ 1/2

= 3/2 ÷ 1/2

= 3/2 × 2/1

= 3

They can make 3 ornaments

8 0
2 years ago
Para reunir dinero para su gira de estudios , los alumnos de un curso deciden vender números de una rifa que se encuentran numer
QveST [7]

Respuesta:

0.53

Explicación:

Para calcular la posibilidad del evento A: "ganar la rifa comprando todos los números múltiplos de 3 o 5", debemos usar la siguiente fórmula.

P(A) = casos favorables / casos posibles

Evaluemos primero todos los casos que son múltiplos de 3, entre 1 y 100: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99. En total son 33.

Ahora, evaluemos todos los casos que son múltiplos de 3, entre 1 y 100: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. En total son 20.

El número total de casos favorables es 33 + 20 = 53.

El número de casos posibles es el total de números de 1 a 100, es decir 100.

Luego P(A) = 53/100 = 0.53.

7 0
2 years ago
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