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Sidana [21]
3 years ago
6

Need help on b ASAP HELP

Mathematics
1 answer:
maxonik [38]3 years ago
3 0
9w^3 ...........................
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Which of the following can be determined about events A and C from the table.
kenny6666 [7]
The correct answer for this question is this one: "<span>C. P(C | A) = 0.75, P(C)=0.75 the events are not independent." 

</span><span>The statement that can be determined about events A and C from the table is that </span><span>P(C | A) = 0.75, P(C)=0.75 the events are not independent. Hope this helps answer your question.
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5 0
3 years ago
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The value of the computer decrease exceptionally throughout the five-year period. What was the average annual amount of decrease
Gemiola [76]

Answer:

Step-by-step explanation:

C grd. Gr

4 0
2 years ago
How many positive integer factors of 100! are multiples of 2 or 3?
elena-s [515]

Positive integer factors of 100:

1, 2, 4, 5, 10, 20, 25, 50, 100

Now find if they are a multiple of 2 or 3.

2, 4, 10, 20, 50, 100

That is 6 factors.

3 0
3 years ago
Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
X2 +x − 4 = 0
Rus_ich [418]

Answer:

  • a=1
  • b=1
  • c=-4
  • x =  (-1 ±√17)/2

Step-by-step explanation:

The coefficient of x^2 is "a". That is 1.

The coefficient of x is "b". That is 1.

The constant term is "c". That is -4.

The values of a, b, and c are 1, 1, and -4, respectively.

_____

The solution is ...

  x = (-b ±√(b^2-4ac))/(2a)

Filling in the values of a, b, and c, this is ...

  x = (-1 ±√(1^2 -4·1·(-4)))/(2·1)

  x = (-1 ±√17)/2

3 0
3 years ago
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