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Lemur [1.5K]
3 years ago
11

Sara has a gift box with a volume of 64 in3. The length, width, and height of the box are all y inches. Sara needs to determine

if a stack of 6-inch-by-6-inch note papers will fit on the bottom of the box. Complete the sentences below by choosing the correct responses from the drop-down menus. Sara can determine the value of y by finding of the volume. The value of y is inches. Therefore, the note papers fit on the bottom of the gift box.
Mathematics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

No, the note papers will not fit the gift box.

Step-by-step explanation:

Given that:

Volume of the gift box = 64 in^3

Length, width and height of the box all are equal to y inches.

Dimensions of note papers = 6 inch by 6 inch

To determine:

Whether the note papers will fit the bottom of the box?

Solution:

Here, we need to find the dimensions of the gift box first.

If the dimensions of gift box are greater than or equal to the dimensions of the note papers, only then the papers will fit into the box.

As per the given question statement, the gift box is in the form of a cube.

Formula for volume of a cube is given as:

Volume = (Side)^3

Side is given equal to y inches.

64 = y^3\\\Rightarrow y = 4\ in

Dimensions of the box are lesser than that of the note papers.

Therefore, <em>the note papers will not fit into the gift box</em>.

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The number of ways to deal each hand of poker is

1) 10200 possibilities

2) 5108 possibilities

3) 40 possibilities

4) 624 possibilities

5) 123552 possibilities

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8) 267696 possibilities

Step-by-step explanation:

Straigth:

The Straight can start from 10 different positions: from an A, from a 2, 3, 4, 5, 6, 7, 8, 9 or from a 10 (if it starts from a 10, it ends in an A).

Given one starting position, we have 4 posibilities depending on the suit for each number, but we need to substract the 4 possible straights with the same suit. Hence, for each starting position there are 4⁵ - 4 possibilities. This means that we have 10 * (4⁵-4) = 10200 possibilities for a straight.

Flush:

We have 4 suits; each suit has 13 cards, so for each suit we have as many flushes as combinations of 5 cards from their group of 13. This is equivalent to the total number of ways to select 5 elements from a set of 13, in other words, the combinatorial number of 13 with 5 {13 \choose 5} .  However we need to remove any possible for a straight in a flush, thus, for each suit, we need to remove 10 possibilities (the 10 possible starting positions for a straight flush). Multiplying for the 4 suits this gives us

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We have 4 suits and 10 possible ways for each suit to start a straight flush. The suit and the starting position determines the straight flush (for example, the straight flush starting in 3 of hearts is 3 of hearts, 4 of hearts, 5 of hearts, 6 of hearts and 7 of hearts. This gives us 4*10 = 40 possibilities for a straight flush.

4 of a kind:

We can identify a 4 of a kind with the number/letter that is 4 times and the remaining card. We have 13 ways to pick the number/letter, and 52-4 = 48 possibilities for the remaining card. That gives us 48*13 = 624 possibilities for a 4 of a kind.

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