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Black_prince [1.1K]
3 years ago
8

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly

selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same
Mathematics
1 answer:
scZoUnD [109]3 years ago
8 0

Answer:

1/3

Step-by-step explanation:

Alice can select any ball from her bag and it goes into Bob's bag. Then bob has 6 balls including two of one color.

Suppose they are {A,B,B,C,D,E}

Now Bob must draw one of the two balls of the same color (B) in order to have the same content of the two bags.

P(bags same content)= 2/6= 1/3

The probability that the bags content are the same= 1/3

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What best describes the asymptote of an exponential function of the form f(x)=b^x?
horrorfan [7]

It depends on the value of b


Exponential functions are defined for positive values of the base [ŧex] b [/tex]. Also, they're not defined if b=1, or at least let's say that this is a trivial case, since it is the constant function 1.


Case 0<b<1:

If the base sits between 0 and 1, the exponential function is constantly decreasing. The explanation is quite intuitive, assume for the sake of simpleness that b is rational. If it sits between 0 and 1, it means that it can be written as a fraction p/q, with p<q. So, if you give a large exponent to b, you obtain


\frac{p^x}{q^x},\quad p^x \ll q^x \text{ as } x \to \infty


On the other hand, if you consider negative exponents, you switch numerator and denominator and then raise to the same exponent the fraction q/p, which gets larger and larger.


So, if 0<b<1, we have


\lim_{x \to -\infty} b^x = \infty,\qquad \lim_{x \to \infty} b^x = 0


and thus 0 is a horizontal asymptote as x tends to (positive) infinity.


Case b>1:

This case is very similar, except all roles are inverted. Now you start with a fraction p/q where p>q. So, with positive, large exponents you get


\frac{p^x}{q^x},\quad p^x \gg q^x \text{ as } x \to \infty


And as before, negative exponents switch numerator and denominator, so the fraction becomes q/p and thus you have


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6 0
3 years ago
Read 2 more answers
Which expressions are equivalent to 35 + 30s - 45t?
stiv31 [10]
We have that
<span>35 + 30s - 45t

case A) 
</span>7⋅(5+30s−45t)---> 35+210s-315t-----> is not <span>equivalent 

case B)
</span>5(7+6s−9t)----> 35+30s-45t------> is equivalent to 35 + 30s - 45t

case C)
(−35−30s+45t)×(−1)----> 35+30s-45t---> is equivalent to 35 + 30s - 45t

case D)
 10×(3.5+3s−4.5)----> 35+30s-45---->  is not  equivalent 

case E) 
( 35/2 - 15s + 45/2t) ⋅ (-2)--->-35+30s-45t---> is not  equivalent 

the answer is
<span>B. 5(7+6s−9t)
C. (−35−30s+45t)×(−1)</span>
3 0
3 years ago
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A function is defined by f (x) = 5 (2 minus x). What is f(–1)? –5 5 15 53
Helen [10]

Answer:

f(-1) = 15

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Functions
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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = 5(2 - x)

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em> [Function f(x)]:                                                                           f(-1) = 5(2 - -1)
  2. (Parenthesis) Subtract:                                                                                     f(-1) = 5(3)
  3. Multiply:                                                                                                             f(-1) = 15
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3 years ago
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