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UNO [17]
3 years ago
11

En una caja de muchos bombones hay hasta tres sabores de ellos. ¿Cuántos se deben extraer como mínimo al azar para obtener con s

eguridad 5 bombones de un mismo sabor?
Mathematics
1 answer:
avanturin [10]3 years ago
3 0

Answer:

At least 13 chocolates must be removed

Step-by-step explanation:

If there are three flavors, the probability of drawing 1 would be: 1

1/3 = 0.333

Which means, that every 3 attempts, theoretically you should do 1 of each, but how they ask for 5 chocolates of each would be:

3 * 5 = 15

At least 15 chocolates must be extracted to theoretically guarantee 5 chocolates each, but how we are interested in knowing is that a single flavor has 5 chocolates, so we discard the last two chocolates that represent the other two flavors

Therefore, for there to be safely 5 chocolates of the same flavor, at least 13 chocolates must be removed.

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Let \mathbf A=\mathbf I be the identity matrix. Then pick whatever matrix you like for \mathbf B.
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Giovanna, Frances, and their dad each carried a 10 pound bag of soil. After putting soil into the first flower bed, Giovanna's b
KIM [24]
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then add and convert to ounces.

ignore my reply it is wrong.
5 0
3 years ago
To rent a jet ski the Delta Water company charges a flat fee of $25 plus $7.50 per hour. If you rented a jet ski for a day of fu
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5 0
3 years ago
Read 2 more answers
Which table of values is correct for the equation y = 5(2)x
Tems11 [23]

Answer:

Option D is correct.

x             y

0             5

1              10

2              20

Step-by-step explanation:

Given the equation:  y = 5(2)^x           .....[1]

Here, x is the input variable and y is the output variable.

For x =0

Substitute in equation [1]; we have;

y = 5(2)^{0} = 5 \cdot 1= 5

For x = 1

Substitute in equation [1]; we have;

y = 5(2)^{1} = 5 \cdot 2= 10

For x =2

Substitute in equation [1]; we have;

y = 5(2)^{2} = 5 \cdot 4= 20

Therefore, the table values which is correct for the equation y = 5(2)^x  is;

x             y

0             5

1              10

2              20

5 0
3 years ago
Consider the expression 2√ 3 cos(x)csc(x)+4cos(x)-3csc(x)-2 √ 3 This expression can be represented as the product of the factors
skad [1K]
The answer:
let be A(x) =<span>2√ 3 cos(x)csc(x)+4cos(x)-3csc(x)-2 √ 3

this function can be represented as </span><span>the product of the factors

proof   
</span>2√ 3 cos(x)csc(x)+4cos(x)-3csc(x)-2 √ 3 = 

2√ 3 cos(x)csc(x)+4cos(x)csc(x) / csc(x) -  3csc(x)- 2 √ 3 csc(x) / csc(x)
this method doesn't change nothing inside the function A(x)

so we have 
[ 2√ 3 cos(x) +4cos(x) / csc(x) -  3 - 2 √ 3 / csc(x) ]  . csc(x)  this is a product of two factors, 
[ 2√ 3 cos(x) +4cos(x) / csc(x) -  3 - 2 √ 3 / csc(x) ]  and csc(x) 

for more explanation

A(x) =[ 2√ 3 cos(x) - 3 + (4cos(x) - 2 √ 3 ) / csc(x)  ]  . csc(x)



8 0
3 years ago
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