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nekit [7.7K]
4 years ago
12

PLEASE HELP! Ordinary brass contains 61.6% copper, 2.9% lead, 0.2% tin, and 35.3% zinc. How many pounds of each metal would be i

n 160 lb. of brass?
Mathematics
2 answers:
MArishka [77]4 years ago
6 0
There would be 98.56 lbs of copper, 4.64 lbs of lead, 0.32 lbs of tin, and 56.48 lbs of zinc.

61.6% = 0.616; 160(0.616)=98.56
2.9% = 0.029; 160(0.029) = 4.64
0.2% = 0.002; 160(0.002) = 0.32
35.3% = 0.353; 160(0.353) = 56.48
dolphi86 [110]4 years ago
4 0

Answer:

\text{Amount of copper}=98.56\text{ lb}

\text{Amount of lead}=4.64\text{ lb}

\text{Amount of tin}=0.32\text{ lb}

\text{Amount of zinc}=56.48\text{ lb}

Step-by-step explanation:

We have been given that ordinary brass contains 61.6% copper, 2.9% lead, 0.2% tin, and 35.3% zinc.

The amount of each metal would be given percent of 160 pounds.

\text{Amount of copper}=\frac{61.6}{100}\times 160\text{ lb}

\text{Amount of copper}=0.616\times 160\text{ lb}

\text{Amount of copper}=98.56\text{ lb}

\text{Amount of lead}=\frac{2.9}{100}\times 160\text{ lb}

\text{Amount of lead}=0.029\times 160\text{ lb}

\text{Amount of lead}=4.64\text{ lb}

\text{Amount of tin}=\frac{0.2}{100}\times 160\text{ lb}

\text{Amount of tin}=0.002\times 160\text{ lb}

\text{Amount of tin}=0.32\text{ lb}

\text{Amount of zinc}=\frac{35.3}{100}\times 160\text{ lb}

\text{Amount of zinc}=0.353\times 160\text{ lb}

\text{Amount of zinc}=56.48\text{ lb}

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Rate of collision,

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or, \frac{1.2}{4}

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So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}


                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

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         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}


                      \simeq 0.7408182207 ------------------------------------(2)

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so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}


                      \simeq 0.2222454662 ------------------------------------(4)

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P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

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                                \frac{1}{6} collision per month

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=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

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= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

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