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Burka [1]
3 years ago
10

a jar of cookies contains 18 different types of cookies. Each type is equally likely to be chosen. Based on your simulation, how

many times must a cookie be chosen in order to get each type?
Mathematics
2 answers:
hjlf3 years ago
8 0

Answer:

Number of Cookies in the jar =18={A,B,C.......,...P,Q,R}

It is given that, each type is equally likely to be chosen.

Probability of an event

                 =\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}

Probability of Getting any of the cookies out of 18 cookies

                           =\frac{1}{18}

When number of outcomes increases , probability of getting each type of cookie will be Higher.

So, there are 18 cookies , when number of each type of cookie will be

=18 × 18

=324 cookies ,means each distinct single cookie is available in 18  containers, means that is there is total of 324 containers.

So, out of 324 containers each cookie must be chosen 18 times.

prohojiy [21]3 years ago
4 0
1/18th of a chance each time you draw. But it really depends on how many are in the jar. I'd say 18.
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Simplify (x + 2/ x^2 + 2x -3) / (x + 2/x^2 - x)
rjkz [21]

Answer:

The simplest form is x/(x + 3)

Step-by-step explanation:

* To simplify the rational Expression lets revise the factorization

  of the quadratic expression

*  To factor a quadratic in the form x² ± bx ± c:

- First look at the c term  

# If the c term is a positive number, and its factors are r and s they

  will have the same sign and their sum is b.

#  If the c term is a negative number, then either r or s will be negative

   but not both and their difference is b.

- Second look at the b term.  

# If the c term is positive and the b term is positive, then both r and

  s are positive.  

Ex: x² + 5x + 6 = (x + 3)(x + 2)  

# If the c term is positive and the b term is negative, then both r and s

  are negative.  

Ex:  x² - 5x + 6 = (x -3)(x - 2)

# If the c term is negative and the b term is positive, then the factor

  that is positive will have the greater absolute value. That is, if

  |r| > |s|, then r is positive and s is negative.  

Ex: x² + 5x - 6 = (x + 6)(x - 1)

# If the c term is negative and the b term is negative, then the factor

  that is negative will have the greater absolute value. That is, if

  |r| > |s|, then r is negative and s is positive.

Ex: x² - 5x - 6 = (x - 6)(x + 1)

* Now lets solve the problem

- We have two fractions over each other

- Lets simplify the numerator

∵ The numerator is \frac{x+2}{x^{2}+2x-3}

- Factorize its denominator

∵  The denominator = x² + 2x - 3

- The last term is negative then the two brackets have different signs

∵ 3 = 3 × 1

∵ 3 - 1 = 2

∵ The middle term is +ve

∴ -3 = 3 × -1 ⇒ the greatest is +ve

∴ x² + 2x - 3 = (x + 3)(x - 1)

∴ The numerator = \frac{(x+2)}{(x+3)(x-2)}

- Lets simplify the denominator

∵ The denominator is \frac{x+2}{x^{2}-x}

- Factorize its denominator

∵  The denominator = x² - 2x

- Take x as a common factor and divide each term by x

∵ x² ÷ x = x

∵ -x ÷ x = -1

∴ x² - 2x = x(x - 1)

∴ The denominator = \frac{(x+2)}{x(x-1)}

* Now lets write the fraction as a division

∴ The fraction = \frac{x+2}{(x+3)(x-1)} ÷ \frac{x+2}{x(x-1)}

- Change the sign of division and reverse the fraction after it

∴ The fraction = \frac{(x+2)}{(x+3)(x-1)}*\frac{x(x-1)}{(x+2)}

* Now we can cancel the bracket (x + 2) up with same bracket down

 and cancel bracket (x - 1) up with same bracket down

∴ The simplest form = \frac{x}{x+3}

5 0
3 years ago
Read 2 more answers
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finlep [7]

{  y    ≥   2x   +   1;    y      >       -2x     -   3 } Unbound



Check out the Uploads!!!




Hope that helps!!!!                        : )

Download pdf
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7 0
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\bf ~~~~~~\textit{parabola vertex form}&#10;\\\\&#10;\begin{array}{llll}&#10;y=a(x- h)^2+ k\\\\&#10;x=a(y- k)^2+ h&#10;\end{array}&#10;\qquad\qquad&#10;vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\&#10;-------------------------------\\\\&#10;\stackrel{\textit{zeros\qquad \qquad \qquad }}{&#10;\begin{cases}&#10;x=-4\implies &x+4=0\\&#10;x=2\implies &x-2=0&#10;\end{cases}}&#10;\\\\\\&#10;\stackrel{\textit{factored vertex form}}{y=a(x+4)(x-2)}\qquad (6,10)\textit{ we also know that }&#10;\begin{cases}&#10;x=6\\&#10;y=10&#10;\end{cases}&#10;\\\\\\&#10;10=a(6+4)(6-2)
3 0
3 years ago
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Suppose you have 60 red marbles and 40 blue marbles in a
andrew-mc [135]

Answer:  The required probability is 0.26.

Step-by-step explanation: Given that there are 60 red marbles and 40 blue marbles in a  box 10 marbles are picked without replacement.

We are to find the probability of selecting 6 red marbles.

Since the marbles are picked up without replacement, so it is a situation of combination.

Let S denote the sample space of the experiment of drawing 10 marbles and E denote the event that 6 marbles are red.

So,

n(S)=^{100}C_{10}=\dfrac{100!}{10!(100-10)!}=\dfrac{100!}{10!90!}=17310309456440,\\\\\\n(E)=^{60}C_6\times^{40}C_4=\dfrac{60!}{6!54!}\times\dfrac{40!}{4!36!}=50063860\times91390.

Therefore, the probability of event E is given by

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{50063860\times91390}{17310309456440}=0.26.

Thus, the required probability is 0.26.

6 0
3 years ago
23.96448 to the nearest tenth
Ulleksa [173]
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8 0
3 years ago
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