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2 years ago

What is the area of the following polygon? Explain your process for how you arrived at the final area. Show your work

Mathematics

1 answer:

Reptile [31]2 years ago

7 0

46 because 14+13=46 and cm

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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

$f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0 } \atop {0}; Otherwise} \right.$

The value of constant C can be obtained as:

$\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1$

$\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1$

$50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1$

$50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1$

$50C[0+\frac{1}{0.5} ] =1$

$100C = 1$ ⇒ $C = \frac{1}{100}$

C = 0.01

3 0

2 years ago

Mikes plumbing charges 68 for coming to your house and 17 per hour for labor if I paid him 357, how many hours, h, was mike at m

Roman55 [17]

Answer:

b. $68 + 17h = 357$

f. 17 hours

Step-by-step explanation:

Charges for coming to my house = $68 (this will be the constant of the equation)

Amount charged per hour labor = $17

hours = h

Total amount paid to Mike = 357

The equation that represents this situation can be expressed as:

$68 + 17h = 357$

Solve for h

$68 + 17h - 68 = 357 - 68$ (subtraction property of equality)

$17h = 289$

$\frac{17h}{17} = \frac{289}{17}$ (division property of equality)

$h = 17$

Mike was in my house for 17 hours.

5 0

3 years ago

I don’t understand this at all please help

Deffense [45]

Answer:

You need 10 of those numbers to be whole numbers, decimals, or fractions (THESE ARE RATIONAL)

You need the other 10 of those numbers to radicals, or numbers similar to pi (THESE ARE IRRATIONAL)

Step-by-step explanation:

4 0

3 years ago

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occ

Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0

3 years ago

Frank's car is 12 feet long. Alex's car is 108 inches

kirill115 [55]

Answer:

21 feet

Step-by-step explanation:

108 inches / 12 = 9 feet.

9+12=21 feet

5 0

2 years ago