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3 years ago

55 percent of 120 books are fictional how many are non fictional

Mathematics

2 answers:

anyanavicka [17]3 years ago

8 0

1-.55 = .45
120*.45 = 54 are non-fiction

adelina 88 [10]3 years ago

4 0

55 percent of 120 is 66.

120 - 54 = 66.

66 Books are fictional

So that being said, the other part of your remaining total is 54. 54 books are non-fictional. (The above where I said 120-54). Since 120 - 66 = 54.

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Suppose I ask you to pick any four cards at random from a deck of 52, without replacement, and bet you one dollar that at least

Tatiana [17]

Answer:

a) No, because you have only 33.8% of chances of winning the bet.

b) No, because you have only 44.7% of chances of winning the bet.

Step-by-step explanation:

a) Of the total amount of cards (n=52 cards) there are 12 face cards (3 face cards: Jack, Queen, or King for everyone of the 4 suits: clubs, diamonds, hearts and spades).

The probabiility of losing this bet is the sum of:

- The probability of having a face card in the first turn

- The probability of having a face card in the second turn, having a non-face card in the first turn.

- The probability of having a face card in the third turn, having a non-face card in the previous turns.

- The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

1) The probability of having a face card in the first turn

In this case, the chances are 12 in 52:

$P_1=P(face\, card)=12/52=0.231$

2) The probability of having a face card in the second turn, having a non-face card in the first turn.

In this case, first we have to get a non-face card (there are 40 in the dech of 52), and then, with the rest of the cards (there are 51 left now), getting a face card:

$P_2=P(non\,face\,card)*P(face\,card)=(40/52)*(12/51)=0.769*0.235=0.181$

3) The probability of having a face card in the third turn, having a non-face card in the first and second turn.

In this case, first we have to get two consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_3=(40/52)*(39/51)*(12/50)\\\\P_3=0.769*0.765*0.240=0.141$

4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

In this case, first we have to get three consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_4=(40/52)*(39/51)*(38/50)*(12/49)\\\\P_4=0.769*0.765*0.76*0.245=0.109$

With these four probabilities we can calculate the probability of losing this bet:

$P=P_1+P_2+P_3+P_4=0.231+0.181+0.141+0.109=0.662$

The probability of losing is 66.2%, which is the same as saying you have (1-0.662)=0.338 or 33.8% of winning chances. Losing is more probable than winning, so you should not take the bet.

b) If the bet involves 3 cards, the only difference with a) is that there is no probability of getting the face card in the fourth turn.

We can calculate the probability of losing as the sum of the first probabilities already calculated:

$P=P_1+P_2+P_3=0.231+0.181+0.141=0.553$

There is 55.3% of losing (or 44.7% of winning), so it is still not convenient to bet.

5 0

4 years ago

What is 0.40506100 in scientific notation

Olenka [21]

Answer:

4.0506100 x 10^-1

Step-by-step explanation:

Because all you need to do is move the decimal point to the right once, you write 10^-1 to show that to get back to the original number, you just multiply 4.0506100 by -0.1

7 0

4 years ago

-10(n+30)=1/2(24-14n)

Pavel [41]

N = -104 is the correct answer

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3 years ago

How does a perpendicular bisector justify sss triangle congruence theorem?

Semmy [17]

Intersects a given segment at a 90 degree angle,and passes through the given segment's midpoint.

4 0

3 years ago