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Sidana [21]
3 years ago
10

A pyramid of lcm x 1cm x 1cm cubes is formed. The bottom layer has 1 anrrangement of these cubes. The second layer has a 9 angan

gement. The layer has a 7 x 7 arrangement. The fouth layer has a 5 x 5 arrangement. The fifth layer has a 3 x 3 arrangement. The toplayer is a single cube. The sidtes and tops of the pyamid were painted (not the bottom). How many square centimeters in total were painted?
Mathematics
1 answer:
loris [4]3 years ago
6 0

Answer:

1444 uf FC fun with that one is a little bit ago and now I have to do it again in a bit of the day I will be there in a few days

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4. From a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of
Degger [83]

Answer:

P=0.228

Step-by-step explanation:

We know that from a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of size six is to be selected.

Therefore, we have 34 people.

We calculate the number of possible combinations:

C^{34}_6=\frac{34!}{6!(34-6)!}=1344904\\

Of the 6 professors we choose 2, and of the other 28 people we choose 4.

We calculate the number of favorable combinations:

C_2^6\cdot C_4^{28}=\frac{6!}{2!(6-2)!}\cdot \frac{28!}{4!(28-4)!}=15\cdot 20475=307125\\

Therefore, the probability is:

P=\frac{307125}{1344904}=0.228

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3 years ago
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Three fair dice are rolled, one red, one green and one blue. What is the probability that the upturned faces of the three dice a
r-ruslan [8.4K]

Answer:   \dfrac{5}{9}

Step-by-step explanation:

When we throw a die , Total outcomes =6

When we throw 3 dice , Total outcomes = 6 x 6 x 6 = 216 [by fundamental counting principle]

Given : Three fair dice are rolled, one red, one green and one blue.

Favorable outcomes : When the upturned faces of the three dice are all of different numbers i.e. no repetition of numbers allowed

By Permutations , the number of favorable outcomes = ^6P_3=\dfrac{6!}{(6-3)!}=\dfrac{6!}{3!}=6\times5\times4=120

The probability that the upturned faces of the three dice are all of different numbers = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

=\dfrac{120}{216}=\dfrac{5}{9}

The probability that the upturned faces of the three dice are all of different numbers  is \dfrac{5}{9} .

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Write two different pairs of decimals whose sums are 8.69. One pair should involve regrouping
chubhunter [2.5K]
One pair is 4.345+4.345 (which is the pair that does not involve grouping)
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3 years ago
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