OMG HELP RN I GIVE BRAINLIST

Mathematics

1 answer:

natta225 [31]2 years ago

3 0

Answer: B

The number 3 isn't included because the point isn't included.

The arrow points to the left, representing any number less than 3 to be a solution.

Inequality: x < 3

You might be interested in

a hardware store marks up the price of all tools it gets from the ware house by 18% and then puts them on sale for 20% off what

Sauron [17]

Ksisiszijssjsndbdn 882728338388 Irish

8 0

2 years ago

A student is trying to solve the system of two equations given below:

erma4kov [3.2K]

Your answer is the third one. WHen it comes to system of equations, this will cancel out 4a, therefore isolating the variable. That turns the equation into -2b=-5.

5 0

2 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$