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MaRussiya [10]
3 years ago
5

An alloy of tin is 15% tin and weights 20 pounds. A second allow is 10% tin. How many pounds of the second alloy must be added t

o the first to get a 12% mixture?
30 lb

40 lb

60 lb
Mathematics
1 answer:
Fofino [41]3 years ago
8 0
Let x represent the amount to be added. The total amount of tin will be
  15%·20 + 10%·x = 12%·(20+x)
  (15%-12%)·20 = (12%-10%)·x
  3%·20/2% = x
  30 = x

30 pounds of 10% tin must be added to get a 12% mixture.
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Answer:

Step-by-step explanation:

See attachment.

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Answer:

(c)

"The given statement is true, by definition of length of a vector v, ||v|| = \sqrt{v\bullet v}"

Step-by-step explanation:

(a) v  \bullet v = || v ||^2

That is completely correct Remember that if  v = (x_1,x_2,x_3)\\ then

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Consider an experiment where two 6-sided dice are rolled. We can describe the ordered sample space as below where the first coor
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Answer:

  • E = { (4,1) , (3,2) , (2,3) , (1,4) }
  • P(E)=\frac{1}{9}
  • P(F|E)=\frac{1}{4}

Step-by-step explanation:

Let's start writing the sample space for this experiment :

S= { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }

Let's also define the event E ⇒

E : '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from S ⇒

E = { (4,1) , (3,2) , (2,3) , (1,4) }

In order to calculate P(E) we are going to divide all the cases favorables to E over the total cases from S. We can do this because all 36 of these possible outcomes from S are equally likely. ⇒

P(E)=\frac{4}{36}=\frac{1}{9} ⇒

P(E)=\frac{1}{9}

Finally we are going to define the event F ⇒

F : '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

F = { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) = P(A,B)

We define the conditional probability as

P(A|B)=\frac{P(A,B)}{P(B)} with P(B)>0

We need to find P(F|E) therefore we can apply the conditional probability equation :

P(F|E)=\frac{P(F,E)}{P(E)}   (I)

We calculate P(E)=\frac{1}{9} at the beginning of the question. We only need P(F,E).

Looking at the sets E and F we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

P(F,E)=\frac{1}{36}

Replacing both probabilities calculated in (I) :

P(F|E)=\frac{P(F,E)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{9}}=\frac{1}{4}=0.25

We find out that P(F|E)=\frac{1}{4}=0.25

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Answer:

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Step-by-step explanation:

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