Help pls 11 points! Plss

Please help !! Solve for x: -2 -2/3 -3/2 -3

Kryger [21]

Answer:

x = $\frac{3}{2}$

Step-by-step explanation:

Given 2 intersecting chords, then the product of the parts of one chord is equal to the product of the parts of the other chord, that is

18x = 9 × 3 = 27 ( divide both sides by 18 )

x = $\frac{27}{18}$ = $\frac{3}{2}$

8 0

3 years ago

Two dice are rolled at the same time, what is the likelihood that both dice added together will be greater than or equal to four

topjm [15]

There is a percentage of 8.333333333333333333333%

3 0

3 years ago

What is the remainder in the synthetic division problem

denis-greek [22]

Answer:

C

Step-by-step explanation:

1 | 4 6 - 2

4 10

------------------

4 10 8 ← remainder

3 0

3 years ago

Read 2 more answers

It is about solving equations by substitution. if you can help plzzz

Lyrx [107]

Message me, and i can help more:)

6 0

3 years ago

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

3 years ago