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Dahasolnce [82]
3 years ago
12

EXPERTS ONLY its not 0 or 128 In how many ways can you put seven marbles in different colors into four jars? Note that the jars

may be empty.
Mathematics
1 answer:
KonstantinChe [14]3 years ago
7 0

Answer:  16384

Step-by-step explanation:

Given, Total jars = 4

Total marbles = 7

Since we need to put marbles in 4 different jars, we need to choose a jar each time.

Possible choices for jars =4

Number of time we need to choose = number of marbles

So, by fundamental counting principle, we have

Total ways to put 7 marbles in 4 jars = 4\times4\times 4\times 4\times4\times 4\times 4

=4^7\\\\=16384

Hence, the required number of ways =16384

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108 ?

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4*9 = 36

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3 years ago
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Which of the following statements is needed in order to prove these triangles are congruent by AAS?
creativ13 [48]

Answer:

RQ\cong UT

Step-by-step explanation:

Two triangles are congruent by AAS postulate if two adjacent corresponding angles are congruent and the next adjacent sides to any of the angles is are also congruent. The adjacent sides should not be in between the two congruent angles.

From the triangles RQS and UTV

\angle R\cong \angle U\\\angle S\cong \angle V

The adjacent side to \angle R is RQ and for \angle U is UT.

Similarly, the adjacent side to \angle S is QS and for \angle V is TV.

So, the possible sides that could be congruent by AAS postulate are:

RQ\cong UT or QS\cong TV

So, the correct option is RQ\cong UT

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3 years ago
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What is the simplified form of 1/x-1/y/1/x+1/y
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3 years ago
A closet contains n pairs of shoes. If 2r shoes are chosen at random, (where 2r < n), what is the probability that there will
Pie
We are choosing 2
2
r
shoes. How many ways are there to avoid a pair? The pairs represented in our sample can be chosen in (2)
(
n
2
r
)
ways. From each chosen pair, we can choose the left shoe or the right shoe. There are 22
2
2
r
ways to do this. So of the (22)
(
2
n
2
r
)
equally likely ways to choose 2
2
r
shoes, (2)22
(
n
2
r
)
2
2
r
are "favourable."

Another way: A perhaps more natural way to attack the problem is to imagine choosing the shoes one at a time. The probability that the second shoe chosen does not match the first is 2−22−1
2
n
−
2
2
n
−
1
. Given that this has happened, the probability the next shoe does not match either of the first two is 2−42−2
2
n
−
4
2
n
−
2
. Given that there is no match so far, the probability the next shoe does not match any of the first three is 2−62−3
2
n
−
6
2
n
−
3
. Continue. We get a product, which looks a little nicer if we start it with the term 22
2
n
2
n
. So an answer is
22⋅2−22−1⋅2−42−2⋅2−62−3⋯2−4+22−2+1.
2
n
2
n
⋅
2
n
−
2
2
n
−
1
⋅
2
n
−
4
2
n
−
2
⋅
2
n
−
6
2
n
−
3
⋯
2
n
−
4
r
+
2
2
n
−
2
r
+
1
.
This can be expressed more compactly in various ways.
4 0
3 years ago
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