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blondinia [14]
3 years ago
6

In 2010, the national debt of the United States was about 14 trillion dollars. In 2003 it was about 7 x 10 to the 12 dollars. Ab

out how many times larger was the national debt in 2010 than 2003? THIS IS SCIENTIFIC NOTATION
Mathematics
2 answers:
maxonik [38]3 years ago
7 0
Okay. 10^12 is 1 trillion. 1 trillion * 7 is 7 trillion. 14 trillion / 7 trillion is 2. The national debt in 2010 is two times larger than in 2003. So that basically means that the national debt doubled from 2003 to 2010.
Vsevolod [243]3 years ago
4 0

Answer:  About 2 times.

Step-by-step explanation:

Given : In 2010, the national debt of the United States was about 14 trillion dollars.

We know that 1 trillion =1,000,000,000,000= 10^{12}

then, In 2010, the national debt of the United States was about 14\times10^{12}

dollars.

In 2003 it was about 7\times10^{12} dollars.

Now, the number of times the national debt was larger in 2010 than 2003 :-

\dfrac{ 14\times10^{12}}{7\times10^{12}}\\\\=\dfrac{14}{7}\times10^{12-12}\\\\=2\times10^0=2

Hence, national debt was about 2 times larger in 2010 than 2003 .

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Consider the following game, played with three standard six-sided dice. If the player ends with all three dice showing the same
ddd [48]

Answer:

a

 P(A) =  \frac{1}{36}

b

P(B) = \frac{15}{36}

c

P(U) = \frac{11}{36}

Step-by-step explanation:

From the question we are told that

   The  number of dice is  n  =  3

Generally  for all three dice to show the same number the second and the third dice must have the same outcome  as the outcome of the first dice.

This means that the number of outcome the first die can  have is  6  (6 sides), the number of outcome which the second and the third dice can have is  1 (they must match the first )

So the probability that all three dice show the same number on the first roll is mathematically represented as

      P(A) =  \frac{6}{6} * \frac{1}{6} *  \frac{1}{6}

=>   P(A) =  \frac{1}{36}

Generally for two of the three dice show the same number after the first roll then the number of outcome for  one of the  dice would be is  6 , the number of outcome for another one must be   1(i.e it must match the first ) , the number of outcome for the remaining one must be   5 (i.e it can show any of the remaining  5  sides which the first and second dice are not showing )

Now the number of ways of selecting this 2 dice that show the same number from the 3 dice is mathematically represented as

     N  =  ^3C_2

Here C stands for combination

So

      N  =  ^3C_2  = 6

So the probability that exactly two of the three dice show the same number after the first roll is mathematically represented as  

      P(B) = N   \frac{1}{6} *  \frac{5}{6}

=>  P(B) = \frac{15}{36}

Generally from the question we are told that if two dice match, the player re-rolls the die that does not match.

Now the probability that  the die that did not match the first  time will match the second time is

     P(E ) =  \frac{1}{6}

Generally if that one die does not show the same number in the second round , the probability that it will match  in the third round is

    P(R) =  \frac{5}{6} *  \frac{1}{6}

Generally the probability that he wins (i.e when all three are showing the same number ) and  exactly two is showing the same number is mathematically represented as

       P(K) =  P(E)* P(B) + P(R)* P(B)

=>    P(K) =  \frac{1}{6} * \frac{15}{36}  +  \frac{5}{6} *  \frac{1}{6} * \frac{15}{36}

=>  P(K)= \frac{165}{1296}

So the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll is mathematically represented as  

    P(U) =  \frac{\frac{165}{1296}}{\frac{15}{36} }

=>  P(U) = \frac{11}{36}

     

 

6 0
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