(30)Points Which Statement is true A, B. C. D.

1)) Which fraction is equivalent to ?

Anni [7]

$\frac{2}{6}$

4 0

2 years ago

Someone help pls I’ll mark brainliest if u give a small explanation! :)

marishachu [46]

Answer:

108 inches^2

Step-by-step explanation:

Multiple each side by 1.5

8x1.5=12

6x1.5=9

then to find the area, multiply the sides together.

4 0

2 years ago

Read 2 more answers

What is the solution to -18-6k=6(1+3k)?

Alexxandr [17]

I'm pretty sure that it's k = -1

3 0

2 years ago

The 17 students on the math team want to raise $483.62 to buy practice books they have already raised $218.25 of each students r

sertanlavr [38]

Answer:

Each stuydent must raise $15.61 to reach their goal.

Step-by-step explanation:

Their goal is 483.62, but they already have 218.25 so we subtract,

483.62 - 218.25 = 265.37

Since there are 17 students and they raise the same amount we need to divide by the amount of students to find out how much money each of them need to raise to add up to their goal.

265.37/17 = $15.61

4 0

2 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

2 years ago