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Molodets [167]
3 years ago
14

David was in need for some money so he decided to clean out the water fountain at the mall he found 20 nickels and quarters. The

collection of nickels and quarters Totaled to 2.6 how many quarters did David find
Mathematics
1 answer:
Shkiper50 [21]3 years ago
3 0

Answer:

David find <u>8</u> quarters.

Step-by-step explanation:

Given:

David was in need for some money so he decided to clean out the water fountain at the mall he found 20 nickels and quarters.

The collection of nickels and quarters totaled to 2.6.

Now, to find the quarters David find.

Let the number of nickels be x

And the number of quarters be y

Thus, the number of nickels and quarters he found:

x+y=20.

⇒ x=20-y.........( 1 )

The totaled of nickels and quarters:

0.05x+0.25y=2.6

(As 1 nickel=$0.05 and 1 quarter=$0.25)

Now, putting the equation ( 1 ) in the place of x we get:

0.05(20-y)+0.25y=2.6

⇒ 1-0.05y+0.25y=2.6

⇒ 1+0.20y=2.6

<em>Subtracting both sides by 1 we get:</em>

⇒ 0.20y=1.6

<em>Dividing both sides by 0.20 we get:</em>

⇒ y = 8.

So, the quarters = 8.

Therefore, David find 8 quarters.

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Step-by-step explanation:

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3 years ago
If you roll two six-faced dice together, you will get 36 possible outcomes. 4 pts 1. List all possible outcomes of the experimen
bezimeni [28]

Given:

Two dice are rolled together.

Total number of possible outcomes.

To find:

The list of total possible outcomes.

The probability of getting a sum of 11 in these outcomes.

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The probability of getting a sum of 13 or more.

Solution:

If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is

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)}

Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3

The probability of getting a sum of 11 in these outcomes is

P(\text{sum=11})=\dfrac{3}{36}

P(\text{sum=11})=\dfrac{1}{12}

Therefore, the probability of getting a sum of 11 in these outcomes is \dfrac{1}{12}.

Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6

The probability of getting a sum less than or equal to 4 is

P({sum\leq 4})=\dfrac{6}{36}

P({sum\leq 4})=\dfrac{1}{6}

Therefore, the probability of getting a sum less than or equal to 4 is \dfrac{1}{6}.

Sum of 13 or more = empty set because maximum sum is 12.

The probability of getting a sum of 13 or more is

P(sum\geq 13)=\dfrac{0}{36}

P({sum\geq 13})=0

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Step-by-step explanation:

The x-intercepts are x = -1 and x = 5, so:

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The vertex is (2, -3), so:

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y = 1/3 (x + 1) (x − 5)

Simplifying:

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Answer:

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Step-by-step explanation:

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