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ELEN [110]
3 years ago
5

I need help solving this

Mathematics
1 answer:
Stolb23 [73]3 years ago
4 0

Step-by-step explanation:

So in the equation given, y = 2x - 3

you substitute x for whats given in the table in the x column.

Example

In the graph the first number under the x colum is -1.

y = 2x - 3 in the equation you take out x and put -1.

So now the equation becomes y = 2 × -1 - 3.

Using bedmas to solve the question you should get -5

Which now means y = -5

To plot the point now x would be -1 and y would be -5 (-1, -5)

Same thing for the second number in the x Column.

y = 2 × 1 - 3 which equals -1

To plot it

x = 1 y = -1. (1, -1)

And for the last number 3.

Agai. You substitute x for 3 which makes the equation y = 2 × 3 - 3

this gives you 3 and to plot it

x would be 3 and y would be 3

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At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
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(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.

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(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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