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Nitella [24]
2 years ago
14

Can anyone help me with this

Mathematics
1 answer:
Aleksandr-060686 [28]2 years ago
8 0

Answer:the 3rd one

Step-by-step explanation:

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4(3x^2y^4)^3 / (2x^3y^5)^4
tatyana61 [14]

Solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Step-by-step explanation:

We need to Solve the expression: \frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

Solving:

\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}

Applying exponent rule: \frac{x^a}{x^b}=x^{a-b}

=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}

Another exponent rule says: x^{-a}=\frac{1}{x^a}=

=\frac{27}{4x^{6}y^{8}}

So, solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Keywords: Solving Exponents  

Learn more about Solving Exponents at:

  • brainly.com/question/13174260
  • brainly.com/question/13174254
  • brainly.com/question/13174259

#learnwithBrainly

8 0
3 years ago
At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

7 0
4 years ago
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