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charle [14.2K]
3 years ago
8

How many different ways are there to arrange the letters A, B, A, and B? 4 6 8 16

Mathematics
1 answer:
andriy [413]3 years ago
6 0

There are 6 different possible arrangements of letters A, B, A, B.

<u>Solution: </u>

Need to determine different ways to range letters A, B, A and B.

Using the theorem which says that the number of permutation of n alphabets, where p_1 number of alphabets of one kind and p_2 is number of alphabets of second kind is given by following formula.

Number of possible arrangements =\frac{n !}{p_{1} ! \times p_{2} !} \rightarrow(1)

In our case total number of alphabets = n = 4

Number of letter A = p_1 = 2

Number of letter B = p_2 = 2

Using (1), we get

Number of possible arrangements of A, B, A, B =\frac{4 !}{2 ! \times 2 !}=\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}=3 \times 2=6

Hence there are 6 different possible arrangements of letters A, B, A, B.

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Question
wlad13 [49]

Answer:

No Solutions

Step-by-step explanation:

First to complete the square, it will need to look something like

a^2+2ab+b^2

so in order to get that we can see that 4n^2+8n seems to be missing a piece.

a is the square root of 4n^2 which is 2n.

8n/2n=4

4/2=2

b=2

so now add 2 to both sides of the equation so now it's

4n^2+8n+4=-13

rewrite as (2n+2)^2=-13

Now you see a negative number as the result of a exponent.

A number squared will never be negative.

Therefore the answer is

No Solutions

7 0
2 years ago
A machine is used to fill plastic bottles with bleach. A sample of 18 bottles had a mean fill volume of 2.007 L and a standard d
abruzzese [7]

Answer:

The 99% confidence interval for the difference between the mean fill volumes at the two locations is;

-0.1175665 L < μ₁ - μ₂ < 0.1295665 L

Step-by-step explanation:

The number of bottles in the sample at the first location, n₁ = 18 bottles

The mean fill volume, \bar{x}_{1} = 2.007 L

The standard deviation, σ₁ = 0.010 L

The number of bottles in the sample at the second location, n₂ = 10 bottles

The mean fill volume, \bar{x}_{2} = 2.001 L

The standard deviation, σ₂ = 0.012 L

The nature of the variance of the two samples = Equal variance

The confidence interval of the statistics, C = 99%

The difference between the mean

\mu_1 - \mu_2 = \left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2} \times \sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}+\dfrac{\sigma _{2}^{2}}{n_{2}}}

(1 - C)/2 = (1 - 0.99)/2 = 0.005, the degrees of freedom, f = n₁ - 1 = 10 - 1 = 9

∴ t_{\alpha /2} = 3.25

Therefore, we have;

\mu_1 - \mu_2 = \left (2.007- 2.001  \right )\pm 3.25 \times \sqrt{\dfrac{0.01^{2}}{18}+\dfrac{0.12^{2}}{10}}

Therefore, we have the difference of the two means given as follows;

-0.1175665 L < μ₁ - μ₂ < 0.1295665 L

8 0
3 years ago
Which of the following is equal to the expression above
EleoNora [17]

Answer:

15 \sqrt[3]{2}

Step-by-step explanation:

{(27 \times 250)}^{ \frac{1}{3} }  =  {(27 \times 125 \times 2)}^{ \frac{1}{3} }  \\  =  {27}^{ \frac{1}{3} }  \times  {125}^{ \frac{1}{3} }  \times  {2}^{ \frac{1}{3} }  \\  =  \sqrt[ 3]{27}  \times  \sqrt[3]{125}  \times  \sqrt[3]{2}  \\  =  \sqrt[3]{ {3}^{3} }  \times  \sqrt[3]{ {5}^{3} }  \times  \sqrt[3]{2}  \\  = 3 \times 5 \times  \sqrt[3]{2}  \\  = 15 \sqrt[3]{2}

4 0
3 years ago
25% of z is 150. Find the value of z.
Nonamiya [84]
The answer is 600
25 x 4 =100
150 x 4 = 600
checking work
.25 x 600= 150
4 0
2 years ago
Read 2 more answers
What is an a priori proposition? an a posteriori proposition?
solniwko [45]
A priori means as things were in the beginning or at the start.
a posteriori means as things are after any event.
4 0
3 years ago
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