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kotegsom [21]
3 years ago
13

A shipping box is 10 1/2 in wide, 16 1/2 in long, and 14 1/2 in high. What is the volume of the box in cubic inches?

Mathematics
2 answers:
ollegr [7]3 years ago
5 0

Answer:

2512.125

Step-by-step explanation:

multiply all three numbers together  

IrinaVladis [17]3 years ago
5 0

The volume of the box is 2512.125 cubic inches.

<u>SOLUTION:</u>

Given that, a shipping box is 10 \frac{1}{2} \text{ in wide}, 16 \frac{1}{2} \text{ in long, and } 14 \frac{1}{2} in high.  

We have to find the volume of the box in cubic inches.

Now, we know that, \text { volume of box }=\text { length } \times \text { width } \times \text { height }

\text { Volume of the box }=10 \frac{1}{2} \text { in } \times 16 \frac{1}{2} \text { in } \times 14 \frac{1}{2} \text { in }

Here 10\frac{1}{2} =10.5; 16\frac{1}{2} =16.5 \text{ and } 14\frac{1}{2} =14.5

\Rightarrow 10.5\times16.5\times14.5

\Rightarrow 2512.125

The required volume is 2512.125 units

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-11.3 so the answer is B
4 0
2 years ago
II'll mark brainliest who is the first to answer both questions correctly.
Strike441 [17]

Answer:

A) The probability that each player gets an ace, a 2 and a 3 with the order unimportant = (72/1925) = 0.0374

B) The probability of winning the jackpot = (1/1200) = 0.0008333

Step-by-step explanation:

A) There are four Aces, four 2's, and four 3's, forming a set of 12 cards

These cards are to be divided at random between 4 players.

What is the probability that each player gets an ace, a 2 and a 3.

We start with the first player, the probability of these cards for the first player, with order not important (because order isn't important, there are 6 different arrangement of the 3 cards)

6 × (4/12) × (4/11) × (4/10) = (16/55)

Then the second player getting that same order of cards

6 × (3/9) × (3/8) × (3/7) = (9/28)

Third player

6 × (2/6) × (2/5) × (2/4) = (2/5)

Fourth player

6 × (1/3) × (1/2) × (1/1) = 1

Probability that each of the players get different cards is then a multiple of the probabilities obtained above

= (16/55) × (9/28) × (2/5) × 1

= (288/7700) = (72/1925) = 0.0374

B) The concluding part of the B question.

To win the jackpot, the numbers on your ticket must match the three white balls and the SuperBall. (You don't need to match the white balls in order). If you buy a ticket, what is your probability of winning the jackpot?

Probability of wimning the jackpot is a product the probability of getting the 3 white balls correctly (order unimportant) and the probability of picking the right red superball

Probability of picking 3 white balls from 10

First slot, any of the 3 lucky numbers can fill this slot, (3/10)

Second slot, only 2 remaining lucky numbers can fill this slot, (2/9)

Third slot, only 1 remaining lucky number can fill this slot, (1/8)

(3/10) × (2/9) × (1/8) = (1/120)

Probability of picking the right red superball

(1/10)

Probability of winning the jackpot = (1/120) × (1/10) = (1/120) × (1/10) = (1/1200) = 0.0008333

Hope this Helps!!!

5 0
3 years ago
What’s is the answer
77julia77 [94]
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3 years ago
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vlada-n [284]
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3 0
3 years ago
Which term can be added to the list so that greatest common factor of the three terms is 12h^3? 36h^3 12h^6. 6h^3. 12h^2. 30h^4.
Juliette [100K]

Answer:

Last Option 48h^{5}.

Step-by-step explanation:

In this question we will do the factors of given two terms then we will do the common factors of all the terms given in answer to match with.

Common factors of 36h³ = 1×2×2×3×3×h×h×h

Common factors of 12h^{6} = 1×2×2×3×h×h×h×h×h×h

In these two terms greatest common factor Of these two terms is = 1×2×2×3×h×h×h = 12h³

Therefore the third term will be the number which has the greatest common factor = 12h³

So the given terms are

6h³ = 1×2×3×h×h×h

12h² = 1×2×2×3×h×h

30h^{4} = 1×2×3×5×h×h×h×h

48h^{5} = 1×2×2×2×2×3×h×h×h×h×h

Therefore the greatest common factor of the term which matches with 12h³ is 48h^{5}

3 0
3 years ago
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