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Elden [556K]
2 years ago
10

How many 3-coin combinations exist if we can choose from

Mathematics
1 answer:
Alex787 [66]2 years ago
3 0

There are four numbers of 3-coin combinations if we can choose from nickels, dimes, quarters, and half-dollars.

<h3>How to solve probability combinations?</h3>

The coins to select from are nickels, dimes, quarters, and half-dollars;

Thus;

Coins (n) = 4

The number of coin to select is:

Coin (r) = 3

The coin combination is then calculated using:

Combination = ⁴C₃

Apply the combination formula, we have;

Combination = 4

Thus, there are four number 3-coin combinations if we can choose from nickels, dimes, quarters, and half-dollars.

Read more about combinations at; brainly.com/question/4658834

#SPJ1

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agasfer [191]

Answer:

0.545 your welcome :D

Step-by-step explanation:

109÷200

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What is the value of a?
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Answer: In right triangle XZW, the height WY is the geometric mean of segments ZY and XY.

Hence,

4^2=3 x a,

a=16/3 un.

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3 years ago
Donte used mental math to find the exact sum of 386 and 14. Which strategy could Donte have used?
Nostrana [21]

Answer:

380 + 6 + 14

Step-by-step explanation:

The sum of 386 and 14 is 386 + 14 = 400.

To find which stategy he used, we find the results of each sum, and it has to be 400.

380 + 6 + 14

380 + 6 + 14 = 386 + 14 = 400

So this is the strategy that he could have used.

380 + 14 + 14

380 + 14 + 14 = 394 + 14 = 408

Result different of 400, which means that this was not the strategy

386 + 10 + 4 + 14

386 + 10 + 4 + 14 = 386 + 14 + 14 = 408

Result different of 400, which means that this was not the strategy

386 + 14 + 80 + 6

386 + 14 + 80 + 6 = 400 + 86 = 486

Result different of 400, which means that this was not the strategy

3 0
3 years ago
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Round 236 to the nearest tenth
bekas [8.4K]
Is it .236 cuz then you would round it to,.24
7 0
4 years ago
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PLEASE HELP!!!!!!!!!!!!!!
Lelechka [254]

Answer:

Step-by-step explanation:

(8x²-18x+10)/(x²+5)(x-3)

express the expression as a partial fraction:

(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +bx+c/x²+5

both denominator are equal , so require only work with the nominator

(8x²-18x+10)=(x²+5)A+(x-3)(bx+c)

8x²-18x+10= x²A+5A+bx²+cx-3bx-3c

combine like terms:

x²(A+b)+x(-3b+c)+5A-3c

(8x²-18x+10)

looking at the equation

A+b=8

-3b+c=-18

5A-3c=10

solve for A,b and c (system of equation)

A=2 , B=6, and C=0

substitute in the value of A, b and c

(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +(bx+c)/x²+5

(8x²-18x+10)/[(x^2+5)(x-3)] = 2/x-3 + (6x+0)/(x²+5)

(8x²-18x+10)/[(x^2+5)(x-3)] =

<h2>2/(x-3)+6x/x²+5</h2>

(4x+2)/[(x²+4)(x-2)]

(4x+2)/[(x²+4)(x-2)]= A/(x-2) + bx+c/(x²-2)

(4x+2)=a(x²-2)+(bx+c)(x-2)

follow the same step in the previous answer:

the answer is :

<h2>(4x+2)/[(x²+4)(x-2)]= 5/4/(x-2) + (3/2 -5x/4)/(x²+4)</h2>

8 0
3 years ago
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