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Elena-2011 [213]
2 years ago
14

A rectangular sandbox measures 4 feet by 7 feet by 2 feet. How many cubic feet of sand can the sandbox hold?

Mathematics
1 answer:
stiks02 [169]2 years ago
7 0

Answer:

56 cubic feet

Step-by-step explanation:

V = l x w x h

V = 4 x 7 x 2

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How do u find length of a obtuse angle
wolverine [178]

Answer:

An obtuse triangle is a triangle that has a single obtuse angle, which is an angle that measures more than 90 degrees and less than 180 degrees. Obtuse triangles, also referred to as oblique triangles, can be recognized by their having a single significantly larger angle and two smaller angles. Since every triangle has a measurement of 180 degrees, a triangle can only have one obtuse angle. You can calculate an obtuse triangle using the lengths of the triangle's sides.

Square the length of both sides of the triangle that intersect to create the obtuse angle, and add the squares together. For example, if the lengths of the sides measure 3 and 2, then squaring them would result in 9 and 4. Adding the squares together results in 13.

Square the length of the side opposite the obtuse angle. For the example, if the length is 4, then squaring it results in 16.

Step-by-step explanation:

3 0
3 years ago
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A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one
TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are \binom44\binom{48}1=48 possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by \binom{13}1=13. So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing \binom41\binom{48}4=778,320 possible hands. Exactly 2 aces are drawn in \binom42\binom{48}3=103,776 hands. And so on. This gives a total of

\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656

possible hands containing at least 1 ace, and hence B occurs with probability

\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. P(A\cap B)=P(A)P(B). This happens if

  • the hand has 4 aces and 1 non-ace, or
  • the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So A\cap B consists of 96 possible hands, which occurs with probability

\dfrac{96}{\binom{52}5}\approx0.0000369

and so the events A and B are NOT independent.

4 0
3 years ago
84859>84949 is this statement true or false?
lord [1]
The answer to this question is:

False. 84859 is less than 84949
6 0
3 years ago
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Enter an inequality that represents the phrase the sum of 1 and Y is greater than or equal to 4
Margaret [11]
Y+1>_4

sum= addition

greater tan or equal to means it has this symbol > with a line underneath it

5 0
3 years ago
25 Points ! Write a paragraph proof.<br> Given: ∠T and ∠V are right angles.<br> Prove: ∆TUW ∆VWU
vladimir2022 [97]

Answer:

Δ TUW ≅ ΔVWU ⇒ by AAS case

Step-by-step explanation:

* Lets revise the cases of congruent for triangles

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and  

 including angle in the 2nd Δ  

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  

 ≅ 2 angles and the side whose joining them in the 2nd Δ  

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

 and one side in the 2ndΔ

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

 leg of the 2nd right angle Δ

* Lets solve the problem

- There are two triangles TUW and VWU

- ∠T and ∠V are right angles

- LINE TW is parallel to line VU

∵ TW // VU and UW is a transversal

∴ m∠VUW = m∠TWU ⇒ alternate angles (Z shape)

- Now we have in the two triangles two pairs of angle equal each

 other and one common side, so we can use the case AAS

- In Δ TUW and ΔVWU

∵ m∠T = m∠V ⇒ given (right angles)

∵ m∠TWU = m∠VUW ⇒ proved

∵ UW = WU ⇒ (common side in the 2 Δ)

∴ Δ TUW ≅ ΔVWU ⇒ by AAS case

7 0
2 years ago
Read 2 more answers
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