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Crank
3 years ago
15

I have 8 pennies and some quarters, nickels, and dimes. The number of pennies plus nickels equals the number of dimes. I have 3

quarters for every 2 pennies and 3 dimes for every 2 quarters. How many coins are nickels?
Mathematics
2 answers:
Sati [7]3 years ago
7 0

Answer:8 pennies, 12 quarters, 18 dimes, so there are 10 nickels

Step-by-step explanation:

Darina [25.2K]3 years ago
5 0

There are five nickels.

Let <em>w</em> = the number of pennies; <em>x</em> = the number of nickels; <em>y</em> = the number of dimes; <em>z </em>= the number of quarters.

Then, we have three equations with four unknowns:

(1) <em>w</em> + <em>x = y </em>

(2) 3<em>w</em> = 2<em>z</em>

(3) 2<em>y</em> = 3<em>z</em>

However, there is a fourth <em>unstated </em>condition: <em>w</em>, <em>x</em>, y, and <em>z</em> must all be integers.

Assume that z = 6 (to avoid fractions).

From Equation (2), w = 4.

From Equation (3), y = 9.

Insert the values for <em>w</em> and <em>y</em> into equation (1).

4 + <em>x</em> = 9

x = 9 – 4 = 5

Thus, there are 4 pennies, 5 nickels, 9 dimes, and 6 quarters.

<em>Check</em>: 4 pennies + 5 nickels = 9 dimes

6 quarters + 4 pennies = 3 quarters for every 2 pennies

9 dimes + 6 quarters = 3 dimes for every 2 quarters

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A company performed power tests on a set of batteries of the same type. The company 10 determined that the equation y = 100 - 8.
alexgriva [62]
For this case we have the following equation:
 y = 100 - 8.9x
 Where,
 x = the number of hours of use
 For 11 hours of use we have x = 11.
 Substituting:
 y = 100 - 8.9 * (11)
 y = 2.1%
 Answer:
 
the best prediction of the percent of remaining power for a battery after 11 hours of use is:
 
y = 2.1%
7 0
3 years ago
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
3 years ago
Which best explains whether a triangle with side lengths 2in , 5in , 4in. is an acute triangle
lutik1710 [3]
So we know that 3,4 and 5 are pythagorean triplet and these result from a right triangle sides length 
- than 3,4 and 5 mean right triangle so 2,5 and 4 mean acute triangle sure or you can prove it in this way too
- than 3^2 +4^2 =5^2 => a right triangle so than 
2^2 +4^2 < 5^2 => an acute triangle 

hope helped 
3 0
3 years ago
If 2 cards are selected from a standard deck of 52 cards.The first card is placed back in the deck before the 2nd card is drawn.
Mkey [24]

There are 13 cards of each suit.

Picking a heart would be 13/52 which reduces to 1/4

Then replacing the card and picking a club would be the same : 1/4

Picking a heart then a club would be 1/4 x 1/4 = 1/16

Answer: 1/16

7 0
2 years ago
Whats the volume of a cone? Lengths: 9cm, 5cm
abruzzese [7]

Answer:

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Step-by-step explanation:

The volume of a cone is given by the formula ...

  V = (1/3)πr²h

Your cone has r = 5 cm and h = 9 cm. Putting these values into the formula gives ...

  V = (1/3)π(5 cm)²(9 cm) = 75π cm³

3 0
3 years ago
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