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ale4655 [162]
3 years ago
9

Solve the equation: 2/3x+1/4=3

Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0

Answer:

x = 33/8

Step-by-step explanation:

2/3 x + 1/4 = 3

Subtract 1/4 from both sides.

2/3 x = 3 - 1/4

2/3 x = 12/4 - 1/4

2/3 x = 11/4

Multiply both sides by 3/2.

3/2 * 2/3 x = 3/2 * 11/4

x = 33/8

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Which of these number is an integer? A. −4.0 B. 1.50 C. -5/8 D. infinity
NNADVOKAT [17]
<span>−4.0 is an integer.

hope it helps</span>
8 0
3 years ago
Need someones help on this!
Paha777 [63]

Answer:

1st, 3rd, and 4th box.

Step-by-step explanation:

9^5=59049

9^2=81   9^3=729    729*81=59049              yes

9^5=59049   59049x9=531441                     no

9^4=6561    6561*9=59049                           yes

9^2=81      81x81=6561     6561*9=59049     yes

9^3=729   729x729=531441                          no

4 0
3 years ago
Read 2 more answers
Solve y2 4y – 32 = 0 using the zero product property
maxonik [38]
I hope this helps you

5 0
3 years ago
Read 2 more answers
If 20+15=24 <br> 64+13=42<br> Then 11+28=?
Rus_ich [418]
2+0+1+5x3= 24 and
6+4+1+3x3=42 than
<span>1+1+2+8x3=36

Final Answer 36 

:D Welcome</span>
5 0
3 years ago
assume that each of 2000 frogss living near a nuclear power plant is exposed to particles of a certain kind of radiation at an a
aleksandr82 [10.1K]

distribution of the total number of tumors produced in the whole population over a one-year period by this kind of radiation is 0, 1, 2, ..................104000.

let, x denote the total number of hits,

y denotes the number of hits by a particle in a year or an individual,

one per week ⇒ 52 per year

therefore, rate = 52 per year

since it is very rare to hit all individuals, we can use Poisson distribution.

therefore by using the formula p (x = x) = (e^(-rate) × rateˣ) ÷ x! ; x = 0, 1, 2, ............( 2000 × 52)

p (x = 2) = (e⁻⁵² × 52ˣ) ÷ x! ; x = 0, 1, 2, ............104000

now each hit has a harm of probability 10⁻⁵.

so p(total no. of tumors produced over a one year period)

= {e⁻⁵² × 52ˣ} ÷ x! × 10⁻⁵ ; x = 0, 1, 2, ............104000

therefore total no. of tumors = 2000 × (e⁻⁵² × 52^{x.10^{-5} } ) ÷ x!  × xⁿ ; x = 0, 1, 2, ............104000

Know more about Poisson distribution here: brainly.com/question/17280826

#SPJ4

(complete question)

Assume that each of the 2000 individuals living near a nuclear power plant is exposed to particles of a certain kind of radiation at an average rate of one per week. Suppose that each hit by a particle is harmless with a probability of 1 - 10⁻⁵, and produces a tumor with a probability of 10⁻⁵. Find the approximate distribution of the total number of tumors produced in the whole population over a one-year period by this kind of radiation.

4 0
1 year ago
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