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makvit [3.9K]
3 years ago
15

A box contains red marbles and green marbles. Sampling at random from the box five times with replacement, you have drawn a red

marble all five times. What is the probability of drawing a red marble the sixth time
Mathematics
1 answer:
olasank [31]3 years ago
3 0

Answer: The probability of drawing a red marble the sixth time is 1/2

Step-by-step explanation:

Here is the complete question:

A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?

Explanation:

Since the sampling at random from the box containing the marbles is with replacement, that is, after picking a marble, it is replaced before picking another one, the probability of picking a red marble is the same for each sampling. Probability, P(A) is given by the ratio of the number of favourable outcome to the total number of favourable outcome.

From the question,

Number of favourable outcome = number of red marbles =10

Total number of favourable outcome = total number of marbles = 10+10= 20

Hence, probability of drawing a red marble P(R) = 10 ÷ 20

P(R) = 1/2

Since the probability of picking a red marble is the same for each sampling, the probability of picking a red marble the sixth time is 1/2

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If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f
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The quadratic function given by:


f(x)=a(x-h)^2+k, \ \ \ a\neq 0


is in vertex form. The graph of f is a parabola whose axis is the vertical line x=h and whose vertex is the point (h, k). So:


To translate the graph of a function to the right, left, upward or downward we have:

For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)


By knowing this things, we can solve our problem as follows:


FIRST.

  • Translating <em>11 units to the left:</em>

g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2


  • Then translating<em> 5 units down:</em>

g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5


Since the new function is fatter, the factor we need to multiply the term (x+11)^2 <em>must be</em> less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.


<em>Therefore, the right answer is </em><em>b. f(x) = 1/2(x + 11)^2 - 5</em>


SECOND.

  • Translating <em>8 units to the right:</em>

g(x)=f(x-8) \\ \\ \therefore g(x)=(x-8)^2


  • Then translating<em> 1 unit down:</em>

g(x)=f(x)-c \\ \\ \therefore g(x)=(x-8)^2-1


As explained in the previous case, there are two factors 1/3 and 3, so we choose the first one.


<em>Therefore, the right answer is </em><em>a. g(x) = 1/3(x - 8)^2 - 1</em>

7 0
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