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Setler79 [48]
3 years ago
9

I need step by step instructions on how to get the y-intercept: -210 for f(x)=x^3-18x^2+107x-210.

Mathematics
1 answer:
Mrrafil [7]3 years ago
6 0
Y=f(X)
At y-intercept, X=0
f(X)= 0^3-18(0)^3+107(0)-210
= -210
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A cube has side length of 2m. What is the surface area of the cube?
SVEN [57.7K]

Answer:

24 square meters.

Step-by-step explanation:

The surface area of the cube consists of 6 congruent squares with side length of 2 m.

The are of one such square is

2\cdot 2=4\ m^2.

Hence, the surface area of the cube is

6\cdot 4=24\ m^2.

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Answer:

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Step-by-step explanation:

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3 years ago
Classify each function as even, odd, or neither even nor odd.
drek231 [11]

The function

f(x) is even

g(x) is neither even nor odd

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Steps:

for an even function it holds that f(-x) = f(x):

f(-x) = (-1)^6 x^6  - (-1)^4 x^ 4 = x^6 - x^4 = f(x) => f is even

for an odd h(x) it holds that h(-x) = -h(x):

h(-x) = (-1)^5x^5-(-1)^3x^3 = -(x^5-x^3) = -h(x) \implies h(x)\,\, \mbox{even}

It is easy to show that g(x) does not match any of the two possibilities above.


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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
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
3 years ago
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