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leva [86]
3 years ago
8

A bag contains four red cards numbered 1 through 4, four white cards numbered 1 through 4, and four black cards numbered 1 throu

gh 4. You choose a card at random.
Mathematics
1 answer:
Liula [17]3 years ago
3 0

Answer:

there is a 1/12 chance of getting one card of any sort

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.help me ............
il63 [147K]
35=x

180-75=105

X+2x=3x

105=3x

X=35
4 0
3 years ago
Can 0.1/100 be simplified?
FrozenT [24]
No 0.1/100 cannot be simplified
5 0
3 years ago
Read 2 more answers
A Create a linear expression
frez [133]

Answer:

(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB. Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .Step-by-step explanation:

3 0
2 years ago
Which expression is equivalent to 5+5t+3t+?
FrozenT [24]

Answer:

10t+3

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
In a doctor's waiting room, there are 14 seats in a row. Eight people are waiting to be seated. Three of the 8 are in one family
lianna [129]
There are 14 chairs and  8 people to be seated. But among the 8. three will be seated together:

So 5 people and (3) could be considered as 6 entities:

Since the order matters, we have to use permutation:

¹⁴P₆ = (14!)/(14-6)! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆  x 3! 

2,162,160 x 6 = 12,972,960 ways.

Another way to solve this problem is as follow:

5 + (3) people are considered (for the time being) as 6 entities:

The 1st has a choice among 14 ways
The 2nd has a choice among 13 ways
The 3rd has a choice among 12 ways
The 4th has a choice among 11 ways
The 5th has a choice among 10 ways
The 6th has a choice among 9ways

So far there are 14x13x12x11x10x9 = 2,162,160 ways
But the 3 (that formed one group) could seat among themselves in 3!
or 6 ways:

Total number of permutation = 2,162,160 x 6 = 12,972,960

4 0
3 years ago
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