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sashaice [31]
3 years ago
15

The value of a new car decreases by about 15% in the first year. How much will a car be worth after one year if its initial valu

e was $1,500? If you get stuck, consider using diagrams or a table to organize your work.
Mathematics
1 answer:
kicyunya [14]3 years ago
7 0

Answer:

$12,750

Step-by-step explanation: when you do the math you get it your welcome my G

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Y = log3 27

27 = 3^x

3^3 = 27  

so x = 3
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Find the volume of a cylinder made by rotating a rectangle with a width of 2.5 feet and a length of 8 feet as shown in the figur
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<span>the question does not present the picture, but this does not interfere with the resolution</span>

we know that
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157 ft³
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3 years ago
what expression represents the following problem: A glass holds 1/ 4 liters of liquid. How many glasses of liquid are in 5 liter
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20... 5÷ 1/420/ \frac{1}{4}


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4 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.

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4 years ago
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562.20 So the answer is A.
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