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Arada [10]
2 years ago
15

After simplifying, which expressions are equivalent? Select three options.

Mathematics
2 answers:
Wewaii [24]2 years ago
6 0

Answer: Second option, third option and fifth option.

Step-by-step explanation:

In order to solve this exercise it is important to remember the multiplication of signs:

(+)(+)=+\\(-)(-)=+\\(-)(+)=-\\(+)(-)=-

Knowing that, you can distribute the sign and add the like terms. Repeat this procedure in each expression given in the exercise.

Therefore, you get:

a.\ (3.4a - 1.7b) + (2.5a - 3.9b)=3.4a - 1.7b + 2.5a - 3.9b=5.9a-5.6b\\\\b.\ (2.5a + 1.6b) + (3.4a + 4b)=2.5a + 1.6b + 3.4a + 4b=5.9a+5.6b\\\\c.\ (-3.9b + a) + (-1.7b + 4.9a)=-3.9b + a -1.7b + 4.9a=5.9a-5.6b\\\\d.\ -0.4b +(6b - 5.9a)=-0.4b +6b - 5.9a=6b-6.3a\\\\e.\ 5.9a - 5.6b

You can identify that the expressions  (2.5a + 1.6b) + (3.4a + 4b), (-3.9b + a) + (-1.7b + 4.9a) and 5.9a - 5.6b are equivalent after simplifying.

Semmy [17]2 years ago
4 0

Answer:

b, c, e

Step-by-step explanation:

did it good luck

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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:

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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.

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From the equation of motion, we know,

s=ut+\frac{1}{2}at^{2}

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a= gravitational force

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A = -9.8m/s^{2} since the ball is thrown upwards.

Plug the known values into the equation.

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