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vladimir1956 [14]
3 years ago
5

SOLVE ASAP PLEASE!! What is the first step in solving the equation 2/3x+1/3x+2=5 ? Subtract 1/3 from each side of the equation.

Add 2 to each side of the equation. Combine like terms. Multiply each side of the equation by 5.
Mathematics
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

Step-by-step explanation: Not quite...

First, add all the values with x together. 2/3x+1/3x= 1x, or just plain x. So now you’re left with x+2=5. To get x alone, subtract the last thing on the side of the equation with x, which is 2. To even it out, subtract 2 from 5 as well. Now you’re left with x=3

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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          

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(2^8x5^-5x19^0)^-2 x (5^-2/2^3)^4 x 2^28

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