Find the values of x and y. (6x +77)° (2y +41) (4x +43) (5y +48)

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 X = temperature increase.

The random variable X follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of X 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 X 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

2 years ago

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

2 years ago

Is 9760-5,220 more or less then 4,000

givi [52]

It's more than 4,000

7 0

2 years ago

Read 2 more answers

Solution to the inequality 16x−33x<37x+27.

astra-53 [7]

The solution to the inequality is:

$x>\frac{-1}{2}$

Solution:

Given inequality is:

$16x-33x$

We have to find the solution to given inequality

$\mathrm{Add\:similar\:elements:}\:16x-33x=-17x$

$-17x$

$\mathrm{Subtract\:}37x\mathrm{\:from\:both\:sides}$

$-17x-37x$

Simplify the above inequality

$-54x$

$\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}$

Remember that, change the inequality sign if you divide or multiply both sides by a negative number

If you divide or multiply both sides by a positive number,the inequality sign will not change

$\left(-54x\right)\left(-1\right)>27\left(-1\right)\\\\54x>-27$

$\mathrm{Divide\:both\:sides\:by\:}54$

$\frac{54x}{54}>\frac{-27}{54}\\\\x>-\frac{1}{2}$

Thus the solution to inequality is found

3 0

2 years ago

Alan wants to install security cameras in his store, which has dimensions of 30 feet by 30 feet. Each camera can view an area of

FinnZ [79.3K]

30×30=900
900/112.5=8 cameras to cover

112.5×4=450
900-450=450 ft left uncovered

8 0

2 years ago

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