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3 years ago

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Tossing coins imagine tossing a fair coin 3 times. (a) what is the sample space for this chance process? (b) what is the assignm

ent of probabilities to outcomes in this sample space?

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Answer:

(a) what is the sample space for this chance process?

If we toss a coin three times then there are total $2^{3}=8$ outcomes

The sample space associated with the given chance process is:

$S= \left \{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\right \}$

(b) what is the assignment of probabilities to outcomes in this sample space?

Since the given sample space has eight outcomes and we know that a fair coin is tosses three times. Therefore, the probability of all the events mentioned in the given sample space is same. Hence we have:

$Probability = \frac{1}{8}$

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Montano1993 [528]

$0.05p-0.02(5-2p)=0.05(p-2)-0.12$

$0.05p-0.02(5-2p)=0.05(p-2)-0.12:p$

$
0.05p-0.1+0.04p=0.05p-0.1-0.12
 
$ $|Expand \ the \ equation|$

$0.09p-0.1=0.05p-0.1-0.12
 
$

$0.09p-0.1=0.05p-0.22$

$
0.09p-0.1-0.05p=-0.22
 

$ $|Subtract \ 0.05p \ from \ both \ sides|$

$
0.04p-0.1=-0.22
 
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$
0.04p=-0.22+0.1
 
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$0.04p/0.04=-0.12/0.04
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8 0

3 years ago

Assume that a company sold 5.75 million motorcycles and 3.5 million cars in the year 2010. The growth in the sale of motorcycles

OverLord2011 [107]

Answer:

Final answer is approx 6.644 years.

Step-by-step explanation:

Given that a company sold 5.75 million motorcycles and 3.5 million cars in the year 2010. The growth in the sale of motorcycles is 16% every year and that of cars is 25% every year.

So we can use growth formula:

$A=P\left(1+r\right)^t$

Then we get equation for motorcycles and cars as:

$A=5.75\left(1+0.16\right)^t$

$A=3.5\left(1+0.25\right)^t$

Now we need to find about when the sale of cars will be more than the sale of motorcycles. So we get:

$3.5\left(1+0.25\right)^t>5.75\left(1+0.16\right)^t$

$3.5\left(1.25\right)^t>5.75\left(1.16\right)^t$

$3.5\left(1.25\right)^t>5.75\left(1.16\right)^t$

$\frac{\left(1.25\right)^t}{\left(1.16\right)^t}>\frac{5.75}{3.5}$

$\left(\frac{1.25}{1.16}\right)^t>1.64285714286$

$t\cdot\ln\left(\frac{1.25}{1.16}\right)>\ln\left(1.64285714286\right)$

$t>\frac{\ln\left(1.64285714286\right)}{\ln\left(\frac{1.25}{1.16}\right)}$

$t>6.6436473051$

Hence final answer is approx 6.644 years.

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