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inn [45]
3 years ago
13

Derek rolls a die many times and records the number of times he rolls a three. Arrange the following situations in order from th

e situation that
gives the largest difference between relative frequency and actual probability to the situation that gives the smallest difference
Derek rolls the
die 10 times
Derek rolls the
die 30 times
Derek rolls the
die 5 times.
Derek rolls the
die 20 times
Mathematics
2 answers:
mrs_skeptik [129]3 years ago
5 0

The sentence sequence is "BDAC".

<u>Step-by-step explanation</u>:

  • Derek is rolling a die many times and he is recording when he rolls a dice of three. So we have to identify from largest to smallest.
  • The first largest is 30 times. Because of 3*10=30, so he rolls three ten times. So the first option is B.
  • The second largest is 20 times. Derek rolled dice three-six times because of 3*6=18. So the second option is D.
  • The third-largest is 10 times. In this, he rolls three times. Because of 3*3=9. So the option is A.
  • The final option is C. Derek rolls five times. 3*1=3. In this, he rolled only one time. So the sequence is BDAC.                                                                  

Veseljchak [2.6K]3 years ago
3 0

There is NO situation that  gives the largest difference between relative frequency and actual probability to the situation that gives the smallest difference.

Step-by-step explanation:

Event (E) = Denotes getting 3 when rolling a single die.

Here first we find the probability of not getting a 3 from rolling a die

The probability of the event (E) is P(E) =(1/6)  

(No. of favorable outcome/Total no. of possible outcomes).

The probability of not getting 3 is 1-P(E)=1–(1/6)=(5/6). =0.8333

1.Times of rolling a dice  =  10 times

Relative frequency          =   1/6 ^10 = 1.67

Actual probability            =  1 - (5/6)^10. = 1.67

Result                                = No difference

2.Times of rolling a dice  =  30 times  

Relative frequency          =   1/6 ^30 = 5

Actual probability            =  1 - (5/6)^30. = 5

Result                                = No difference

3. Times of rolling a dice  =  5 times

Relative frequency          =   1/6 ^5 = 0.835

Actual probability            =  1 - (5/6)^5. = 0.835

Result                                = No difference  

4. Times of rolling a dice  =  20 times

Relative frequency          =   1/6 ^20 =  3.341

Actual probability            =  1 - (5/6)^20. = 3.341

Result                                =  No difference  

here the times of rolling a dice were different but the dice been a single one having only 6 out comes .,so the outcomes of both relative frequency and actual probability is same in all the cases.,

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**Spam answers will not be tolerated**
Morgarella [4.7K]

Answer:

f'(x)=-\frac{2}{x^\frac{3}{2}}

Step-by-step explanation:

So we have the function:

f(x)=\frac{4}{\sqrt x}

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Therefore, our derivative would be:

\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}

Place the 4 in front:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})

Distribute:

=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}

The numerator will use the difference of two squares. Thus:

=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Simplify the numerator:

=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Both the numerator and denominator have a h. Cancel them:

=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Now, substitute 0 for h. So:

=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})

Simplify:

=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

=4( \frac{-1}{(x)(2\sqrt{x})})

Multiply across:

= \frac{-4}{(2x\sqrt{x})}

Reduce. Change √x to x^(1/2). So:

=-\frac{2}{x(x^{\frac{1}{2}})}

Add the exponents:

=-\frac{2}{x^\frac{3}{2}}

And we're done!

f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}

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Answer:

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3 years ago
PLEASE HELP ME!!!!!!!
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3 years ago
Find an equation, or a set of equations, to describe the set of points that are equidistant from the points p(−9, 0, 0) and q(3,
wariber [46]
X = -3.  
The distance from p(-9, 0, 0) is
 d = sqrt((x+9)^2 + y^2 + z^2) 
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 d = sqrt((x-3)^2 + y^2 + z^2) 
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 sqrt((x+9)^2 + y^2 + z^2) = sqrt((x-3)^2 + y^2 + z^2)
  Square both sides, then simplify
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 x^2 + 18x + 81 + y^2 + z^2 = x^2 - 6x + 9 + y^2 + z^2
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 24x + 81 = 9
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7 0
3 years ago
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