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4 years ago

You roll a die pick a card and find the sum

Mathematics

1 answer:

Kitty [74]4 years ago

8 0

Answer:

1/312

Step-by-step explanation:

I'm guessing this is what you want since sum is adding, but this is a probability question. There are six possibilities or sides on a dice and 52 possibilities or cards in a card deck. The probability of you having a single type of outcome is:

1/6x1/52=1/312

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The value of p so that the expression 8+5n is equivalent to p(4+2.5n).

marissa [1.9K]

Answer:

8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n)sa

Step-by-step explanation: 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n) 8+5n is equivalent to p(4+2.5n)

4 0

3 years ago

Need help please. Please help me.

prisoha [69]

Answer:

Exact Form:

2/45

Decimal Form:

0.04 (4 being repeated)

8 0

4 years ago

A Halloween trick-or-treat group consists of 3 Trolls, 4 Nazguls, 4 Ents, and 5 Pokemon. A committee of 4 is to be picked to rep

balu736 [363]

Answer:

The probability that the committee will consist of 1 from each type of monster is 0.1318 and Suppose that the Pokemon refuse to be on the same committee as a the Trolls.So,the probability that this type of committee is formed is 0.5357

Step-by-step explanation:

No. of Trolls = 3

No. of Nazguls =4

No. of Ents = 4

No. of Pokemon = 5

Total Monsters = 16

We are given that A committee of 4 is to be picked to represent the group at the Monster Bash.

(a) Find the probability that the committee will consist of 1 from each type of monster.

So, the probability that the committee will consist of 1 from each type of monster:

= $\frac{^3C_1 \times ^4C_1 \times ^4C_1 \times ^5C_1}{^{16}C_4}$

= $\frac{\frac{3!}{1!(3-1)!} \times \frac{4!}{1!(4-1)!} \times \frac{4!}{1!(4-1)!}\times \frac{5!}{1!(5-1)!}}{\frac{16!}{4!(16-4)!}}$

= $\frac{\frac{3!}{1!(2)!} \times \frac{4!}{1!(3)!} \times \frac{4!}{1!(3)!}\times \frac{5!}{1!(4)!}}{\frac{16!}{4!(12)!}}$

= $0.1318$

b)Suppose that the Pokemon refuse to be on the same committee as a the Trolls.

So, the probability that this type of committee is formed

= $\frac{(^3C_3 \times ^8C_1)+(^3C_2 \times ^8C_2)+(^3C_1 \times ^8C_3)+(^3C_0 \times ^8C_4)+(^5C_4 \times ^8C_0)+(^5C_3 \times ^8C_1)+(^5C_2 \times ^8C_2)+(^5C_1 \times ^8C_3)}{^{16}C_4}$

= $\frac{(\frac{3!}{3!(3-3)!} \times \frac{8!}{1!(8-1)!})+(\frac{3!}{2!(3-2)!} \times\frac{8!}{2!(8-2)!})+(\frac{3!}{1!(3-1)!} \times\frac{8!}{3!(8-3)!})+(\frac{3!}{0!(3-0)!} \times \frac{8!}{4!(8-4)!})+(\frac{5!}{4!(5-4)!} \times \frac{8!}{0!(8-0)!})+(\frac{5!}{3!(5-3)!} \times \frac{8!}{1!(8-1)!})+(\frac{5!}{2!(5-2)!} \times\frac{8!}{2!(8-2)!})+(\frac{5!}{1!(5-1)!} \times \frac{8!}{3!(8-3)!})}{\frac{16!}{4!(16-4)!}}$

= $0.5357$

Hence The probability that the committee will consist of 1 from each type of monster is 0.1318 and Suppose that the Pokemon refuse to be on the same committee as a the Trolls.So,the probability that this type of committee is formed is 0.5357

4 0

3 years ago

If A = (0,0) and B = (8,2), what is the length of AB

Marina CMI [18]

Answer:

Hope the picture will help you