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borishaifa [10]
2 years ago
11

How many distinct arrangements can you make using the letters in the word experiment

Mathematics
1 answer:
Ratling [72]2 years ago
8 0
E x p e r i m e n t

Total number of letters = 10

Repeated letters =   3 e

Total arrangement from the 10 letters = 10!

Because of the repeated letters =    10! / 3!

                                                       = (10 × 9 × 8 × 7× 6 × 5 × 4 × 3!) / 3!

                                                       = 10 × 9 × 8 × 7× 6 × 5 × 4

                                                       = 604800  

604800 distinct arrangements.
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What is the answer ?
kifflom [539]
The system of inequalities is the following:

i) <span>y ≤ –0.75x
ii)</span><span>y ≤ 3x – 2

since </span>0.75= \frac{75}{100}= \frac{3}{4}, we can write the system again as 

i) y \leq - \frac{3}{4}x
ii) y  \leq 3x-2

Whenever we are asked to sketch the solution of a system of linear  inequalities, we:

1. Draw the lines
2. Color the regions of the inequalities.
3. The solution is the region colored twice.


A.

to draw the line y =- \frac{3}{4}x

consider the points: (-4, 1) and (0, 0), or any 2 other points (x,y) for which y =- \frac{3}{4}x hold.

since we have an "smaller or equal to" inequality, the line is a solid line (not dashed, or dotted).

In order to find out which region of the line to color, consider a point not on the line, for example P(1, 1), which is clearly in the upper region of the line.

For (x, y)=(1, 1) the inequality y  \leq - \frac{3}{4}x, does not hold because 

1 \leq - \frac{3}{4}*1= -\frac{3}{4} is not true,

this means that the solution is the region of the line not containing (1, 1), as shown in picture 1.


B.
similarly, to draw the solution of inequality ii) y ≤ 3x – 2, 

we first draw the line y=3x-2, using the points (0, -2) and (2, 4), or any other 2 points (x,y) for which y=3x-2 holds.

after we draw the line, we can check the point P(1, 7) which clearly is above the line y=3x-2.

for (x, y) = (1, 7), the inequality y ≤ 3x – 2 does not hold

because 7 is not ≤ 3*1-2=1, so the region we color is the one not containing P(1, 7), as shown in picture 2.


The solution of the system is the region colored with both colors, the solid lines included. Check picture 3.

the lines intersect at (0.533, -0.4) because:

–0.75x=3x-2
-0.75x-3x=-2
-3.75x=-2, that is x= -2/(-3.75)=0.533

for x=0.533, y=3x-2=3(0.533)-2=-0.4

Answer: Picture 3, the half-lines included. So the graph is in the 3rd and 4th Quadrants

8 0
3 years ago
Hamburger sells 3 pounds for six dollars if Samantha 510 pounds of hamburger how much will she pay
Lina20 [59]
I believe you meant to say that if Samantha bought 510 pounds of hamburger, how much will she pay?
 
Now from the information we already have, we know that 3 pounds of hamburger will cost 6 dollars. 

The next thing we need to know is how much one pound of hamburger will cost. This will help us calculate  how much the rest will cost.

we now form an equation to assist us.
3 = 6
1= x

(where x represents the unknown price of one pound of hamburger)

Now we must cross-multiply:

3 * x = 6 x 1

3x = 6 

x = 6/3

x = 2

therefore one pound of hamburger costs 2 dollars.

Now, If one pound costs 2 dollars, how about 510 pounds?

We simply multiply:  510 x 2 = 1,020

Therefore  Samantha will pay 1, 020 dollars for 510 pounds of hamburger.






5 0
3 years ago
The cost of 90 kg of rice is 4500. what is the cost of 2 quintal of rice?​
gayaneshka [121]

Answer:

The cost of 2 quintal of rice is $10,000.

Step-by-step explanation:

To determine the cost of 2 quintals of rice, knowing that 90 kilos of said product is worth $ 4,500, it is first necessary to establish the equivalence between quintals and kilograms. In this regard, a quintal is equivalent to 100 kilograms, so 2 quintals are equivalent to 200 kilos.

Now, to determine the cost per kilogram of rice, the following calculation is required:

4,500 / 90 = X

50 = X

Therefore, 1 kilogram of rice costs $ 50. Thus, since 200 x 50 equals 10,000, 2 quintals of rice will cost $ 10,000.

5 0
3 years ago
How many 4-digit numbers are neither multiples of 2 nor multiples of 5?
Gwar [14]

There are 10,000 total four-digit numbers (1000 through 9999).

Multiples of 2 end in 0, 2, 4, 6, and 8. There are 9*10*10*5 = 4500 four-digit multiples of 2.

Multiples of 5 end in 0 or 5. There are 9*10*10*2 = 1800 four-digit multiples of 5.

There is redundancy between the two sets of numbers, namely those that end in 0, which are both multiples of 2 and 5. There are 9*10*10*1 = 900 four-digit multiples of both 2 and 5.

Then there are 4500 + 1800 - 900 = 5400 total four-digit numbers that are either multiples of 2 or 5, which means the remaining 4600 numbers are neither multiples of 2 nor 5.

4 0
3 years ago
Pleaseeee help meeeeeee
Rasek [7]
I think it's the first one :?
5 0
3 years ago
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