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Elina [12.6K]
3 years ago
9

10m^3 of air weigh 13 kg.

Mathematics
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

The weigh of air in the room having volume 38.22 m³ is 49.68 kg  

Step-by-step explanation:

Given as :

Length of room = 4.2 m

Width of room = 3.5 m

Height of room = 2.6 m

So ,  volume of room = Length × Width × Height

Or,   volume of room = 4.2 m × 3.5 m × 2.6 m

∴,     volume of room = 38.22 m³

Now,

∵  10 m³ of air weigh 13 kg

so, 1 m³ of air weigh =  \frac{13}{10} kg

∴    38.22 m³ of air weigh = \frac{13}{10} kg × 38.22 kg

I.e 38.22 m³ of air weigh = 49.68 kg

Hence The weigh of air in the room having volume 38.22 m³ is 49.68 kg  Answer

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Step-by-step explanation:

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A realtor sold a home for ​$341,100 The commission was 4​% of the sale​ price; however, the realtor receives only 60% of the com
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Answer:

The relator recived $8,186.4

Step-by-step explanation:

What you have to do is find the 60% of the commission. Commission is 4% of $341,100 (100%)

First do a cross multiplication to find 4% of $341,100

100% ___ $341,100

4%______x:

x=(4*341,100)/100=13,644

So, the 4% of $341,100 is $13,466

Now you have to find the 60% of $13,466

100% ___ $13,466

60%______x:

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The answer is: $8,186.4

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Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c
GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0

The value of E (X) is 10.

The parameter λ is:

\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10

(a)

Compute the value of P (X > 10) as follows:

P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

P(X

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of <em>x</em> as follows:

P(X

Take natural log on both sides.

ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30

Thus, the value of x is 30.

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