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2 years ago

3

Three letters from the word FREEDOM are selected and arranged in a row. How many different arrangements of the three letters are

obtained, given that repetition of letters is allowed?

2 answers:

mixas84 [53]2 years ago

4 0

There will be 210 ways to arrange it.

Hope it helps! :D

Luden [163]2 years ago

3 0

You can pick 3 letters from the group of 7 letters by combination. C(7,3)=35. Then you you can arrange them in 3!=6 ways. For example you can pick 3 letters for the first place, there left 2 letters, then pick one of the 2 letters for the second place and last letter goes to the third place. In the end you should multiply 35 with 6. Think each choice of picking 3 letters and arrangements as different possibilities. The result should be 35*6=210. Hope it works :)

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a garden plot is to contain 240 sq. ft. if its length is to be 3 times its width,what should its dimensions be?

Nastasia [14]

W^2=80
W SQRT80
W=8.94 ANS.FOR THE WIDTH
L=3*8.94=26.83 ANS.FOR THE LENGTH
PROOF:
240=8.94*26.83
240=240

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2 years ago

Use the Midpoint Formula three times to find the three points that divide the line segment joining (x1, y1) and (x2, y2) into fo

fredd [130]

We are given points $A(x_1,y_1)$ and $B(x_2,y_2)$ .

We first find the midpoint M, of AB, which divides the segment AB into 2 equal parts,

then we find the midpoint N of AM, and midpoint K of MB.

Thus each of the half parts is divided into 2 equal parts. The whole segment is divided into 4 equal parts.

The coordinates of M, N and K are found as follows:

the coordinates of M are: $( \frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})$

the coordinates of N are:

$\displaystyle{( \frac{x_1+\frac{x_1+x_2}{2}}{2} , \frac{y_1+\frac{y_1+y_2}{2}}{2})=( \frac{\frac{2x_1+x_1+x_2}{2}}{2} , \frac{\frac{2y_1+y_1+y_2}{2}}{2})$

$=\displaystyle{(\frac{3x_1+x_2}{4}, \frac{3y_1+y_2}{4})}$

similarly, the coordinates of k are:

$=\displaystyle{(\frac{x_1+3x_2}{4}, \frac{y_1+3y_2}{4})}$

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2 years ago

The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what bounda

blondinia [14]

Answer:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

$\mu = 514 , \sigma =40$

For this case we can use the Chebysev theorem who states that "At least $1 -\frac{1}{k^2}$ of the values lies between $\mu -k\sigma$ and $\mu +k\sigma$ "

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

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2 years ago

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riadik2000 [5.3K]

Answer:

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Isabella- 1 is the starting population. 1+0.2=1.2 is the rate at which it grows each day.

Step-by-step explanation:

Marco's equation should be $10(2^{d})$ since the bacteria double each day. 10 is the starting value of the population. 2 is the growth rate of "double each day" with d as an exponent. This will double each day because:

Day 1 is $2^{1}=2$

Day 2 is $2^{2}=4$

Day 3 is $2^{3}=8$

Day 4 is $2^{4}=16$

You'll notice the value doubles each day.

Isabella has a different equation because her population increases by a percentage. We use the simple interest formula to calculate the bacteria's daily increase or interest.

1(1+0.2)d

1 is the starting population.

1+0.2=1.2 is the rate at which it grows each day.

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marin [14]

The zeros are 3 and -5. Hope this helps!

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2 years ago