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timofeeve [1]
3 years ago
9

Graph of −x − 2y = 6

Mathematics
2 answers:
nlexa [21]3 years ago
5 0

Answer:

Step-by-step explanation:

Andreyy893 years ago
3 0

Answer:

y = 1/2 x - 3

Step-by-step explanation:

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40. Divide 81 by 41. 41 goes into 81 once with a remainder of 40
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Find the discount rate
kifflom [539]

we'd do the same as before on this one as well.

if we take 27.99 to be the 100%, what is 12 off of it in percentage?

\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 27.99&100\\ 12&x \end{array}\implies \cfrac{27.99}{12}=\cfrac{100}{x}\implies 27.99x=1200 \\\\\\ x=\cfrac{1200}{27.99}\implies x\approx 42.87

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What is the value of 4 x 3 divided by (12 divided by2)^2
Digiron [165]

Answer:0.333

Step-by-step explanation:

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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
Ethan opened a savings account that compounds interest yearly. The value of Ethan's account is given by the equation y = 2000(1.
ivanzaharov [21]

Answer:

D

Step-by-step explanation:

hope this helps but I'm not sure if it's right sorry if wrong

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