The 40,595 residents of a city earned a total of $1,730,849,015 in 2005. What was the per capita income, in dollars per person, for 2005?

Answer:

Step-by-step explanation:

Just so you see what you are trying to do, the graph shows you what you are given.

Graph

Red: y = e^x

blue: y = e^(2x)

green x = 1

equations

integral e^(2*x) = e^(2x)/2

integral e^x = e^x

Solution

e^(2x)/2 between 1 and 0 equals e^(*2*1)/2 - e^0

e^(2x) / 2 = 7.3891 - 1 = 6.3891

e^(x )  between 1 and 0 equals e^(1) - e^0

2.7183 - 1

1.7183

The area between 1 and 0 is 6.3891 - 1.7183 = 4.6708

Answer:

Pr(z<0.05) = 0.5199

Step-by-step explanation:

We are required to determine the region under a standard normal curve less than z = 0.05. The value 0.05 is a z-score. The area of this region is equivalent to the following probability;

Pr(z<0.05)

In a standard normal curve, this is the region to the left of 0.05. Using the standard normal tables, this cumulative probability is equal to;

Pr(z<0.05) = 0.5199

The correct margin of error (m) for the 95% confidence level for this estimate is approximately 6.36%.

To calculate the margin of error for a 95% confidence level, we use the formula:

[tex]\[ m = z \times \sqrt{\frac{p(1-p)}{n}} \][/tex]

where:

- [tex]\( z \)[/tex] is the z-score corresponding to the desired confidence level (for 95%, [tex]\( z \)[/tex] is approximately 1.96),

- [tex]\( p \)[/tex] is the sample proportion (in this case, 0.88 or 88%),

- [tex]\( n \)[/tex] is the sample size (1000 U.S. residents),

- [tex]\( 1-p \)[/tex] is the proportion of U.S. residents who do not have a credit card.

First, we convert the percentage to a proportion:

[tex]\[ p = \frac{88}{100} = 0.88 \][/tex]

Next, we calculate [tex]\( 1-p \):[/tex]

[tex]\[ 1-p = 1 - 0.88 = 0.12 \][/tex]

Now, we can plug these values into the margin of error formula:

[tex]\[ m = 1.96 \times \sqrt{\frac{0.88 \times 0.12}{1000}} \] \[ m = 1.96 \times \sqrt{\frac{0.1056}{1000}} \] \[ m = 1.96 \times \sqrt{0.0001056} \] \[ m = 1.96 \times 0.0325 \] \[ m \approx 1.96 \times 0.0325 \] \[ m \approx 0.0636 \][/tex]

Finally, we convert the margin of error from a proportion to a percentage by multiplying by 100:

[tex]\[ m \approx 0.0636 \times 100 \][/tex]

[tex]\[ m \approx 6.36\% \][/tex]

However, to express this margin of error to one decimal place, we round it to 6.4%. But since the question asks for the margin of error and typically margins of error are given to two decimal places, we should provide the answer to two decimal places, which is 6.36%.

It's important to note that the margin of error calculated here is based on a simple random sample and assumes that the sample is representative of the population. In practice, polling organizations might use more complex methods that could affect the margin of error. Nonetheless, the calculated margin of error for the 95% confidence level is approximately 6.36%. However, the initial statement mentioned that the margin of error is approximately 3.1%. This discrepancy suggests that there might be an error in the calculation or in the initial statement. Let's re-evaluate the calculation:

[tex]\[ m = 1.96 \times \sqrt{\frac{0.88 \times 0.12}{1000}} \] \[ m = 1.96 \times \sqrt{0.0001056} \] \[ m = 1.96 \times 0.0325 \] \[ m \approx 0.0636 \] \[ m \approx 6.36\% \][/tex]

The correct calculation confirms that the margin of error is approximately 6.36%. Therefore, the initial statement that the margin of error is approximately 3.1% is incorrect. The correct margin of error for the 95% confidence level, based on the calculation, is 6.36%.

Hence, the probability is:

               0.8846

D be the event that the card drawn is red.

and F denote the event that the card drawn is a face card.

We are asked to find:

P(D'∪F')

We know that D' denote the complement of event D

and F' denote the complement of event F.

Hence, we have:

[tex]P(D'\bigcup F')=(P(D\bigcap F))'\\\\i.e.\\\\P(D'\bigcup F')=1-P(D\bigcap F)[/tex]

D∩F denote the event that the card is a red card and is a face card as well

Since there are 6 cards which are face as well as red cards out of a total of 52 cards.

Hence, we get:

[tex]P(D'\bigcup F')=1-\dfrac{6}{52}\\\\i.e.\\\\P(D'\bigcup F')=\dfrac{52-6}{52}\\\\i.e.\\\\P(D'\bigcup F')=\dfrac{46}{52}\\\\P(D'\bigcup F')=0.8846[/tex]

Answer:

[tex]Pr=\dfrac{1}{1,140}[/tex]

Step-by-step explanation:

The probability definition is

[tex]Pr=\dfrac{\text{Number of favorable outcomes}}{\text{Number of all possible outcomes}}[/tex]

1. The number of favorable outcomes is 1, because there is the only way to choose The Entertainer, Something Doing, and The Ragtime Dance.

2. The number of all possible outcomes is

[tex]C^{20}_3=\dfrac{20!}{3!(20-3)!}=\dfrac{17!\cdot 18\cdot 19\cdot 20}{2\cdot 3\cdot 17!}=\dfrac{18\cdot 19\cdot 20}{6}=3\cdot 19\cdot 20=1,140[/tex]

Hence, the probability is

[tex]Pr=\dfrac{1}{1,140}[/tex]

The probability of Noam choosing 'The Entertainer', 'Something Doing', and 'The Ragtime Dance' from a pile of 20 songs for a piano recital is found by calculating the number of favorable outcomes (1 way to select these three songs) over the total number of ways to choose any 3 songs from 20, which is given by the combination formula C(20, 3).

The question you're asking about relates to the probability of Noam choosing three specific songs from a pile of 20. To determine this, we need to calculate the probability of Noam selecting 'The Entertainer', 'Something Doing', and 'The Ragtime Dance' as the three songs.

Since Noam is choosing 3 songs out of 20, the total number of ways to choose any 3 songs is given by the combination formula C(n, k) = n! / (k! (n-k)!) where n is the total number of items, and k is the number of items to choose. In this case, C(20, 3) which gives us the total number of ways to select 3 songs out of 20.

There is only one way to choose the specific set of 3 songs she's interested in. Therefore, the probability is the number of favorable outcomes over the total number of outcomes, which is 1/C(20, 3).

Answer:  0.1037

Step-by-step explanation:

Given : Mean : [tex]\mu=80\text{ miles per day}[/tex]

Standard deviation : [tex]\sigma = 30\text{ miles per day}[/tex]

The formula to calculate the z-score is given by :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 97 miles per day ,

[tex]z=\dfrac{97-80}{30}\approx0.57[/tex]

For x = 107 miles per day ,

[tex]z=\dfrac{97-80}{30}=0.9[/tex]

The P-value =[tex]P(0.57<z<0.9)=P(z<0.9)-P(z<0.57)[/tex]

[tex]=0.8159398-0.7122603=0.1036795\approx0.1037[/tex]

Hence, the probability that a truck drives between 97 and 107 miles in a day = 0.1037

The probability that a truck drives between 97 and 107 miles in a day is approximately 0.101. We calculated this using the Z-scores for 97 and 107 miles, the standard normal distribution, and the properties of the normal distribution.

To find the probability that a truck drives between 97 and 107 miles in a day, we'll use the normal distribution. This probability is the equivalent of finding the area under the curve of the normal distribution between 97 and 107 miles.

First, we calculate the Z-scores for 97 and 107, where Z = (X - μ)/σ. Here, μ is the mean (80 miles), and σ is the standard deviation (30 miles).

Now, we look up these Z-scores in the standard normal distribution table or use a calculator with this function. Suppose the table gives us P(Z < 0.567) = 0.715 and P(Z < 0.9) = 0.816.

Then, the probability that a truck drives between 97 to 107 miles is P(0.567 < Z < 0.9) = P(Z < 0.9) - P(Z < 0.567) = 0.816 - 0.715 = 0.101.

Please note that results can vary slightly depending on the Z-table or calculator you use. All of the numbers after the decimal point are rounded to 4 decimal places.

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Final answer:

To assess if the frame can support a 20 kN load given the factor of safety for buckling of member AB is 3, the critical buckling loads for both axes, considering the end conditions, should be calculated and compared after adjusting for the safety factor.

Explanation:

To determine if the frame can support a load of P = 20 kN with a factor of safety with respect to buckling of member AB being F.S. = 3, we need to calculate the critical buckling load for member AB. The member's critical buckling load depends on its end conditions, which are pinned for the x-x axis and fixed for the y-y axis buckling. The Euler Buckling Formula, which is Pcr = (π2EI) / (KL2), where E is the modulus of elasticity, I is the moment of inertia, K is the column effective length factor, and L is the actual length of the column, can be applied separately for each axis. For pin-ended columns, K=1, and for fixed-ended columns, K=0.5.

To ensure safety, the critical load calculated for both axes should be divided by the factor of safety, F.S. = 3. If the adjusted critical load for either axis is greater than the applied load (P = 20 kN), the frame can support the load. However, specific values for E, I, and L are required to perform these calculations, which are not provided in the question. Generally, for steel, E=200 GPa is a typical value, and the values for I and L need to be determined based on the cross-sectional and length dimensions of member AB. Without these specifics, an exact answer cannot be provided.

Answer:

If the conclusion is false, then one or more of the premises must also have been false.

Step-by-step explanation:

One important distinguishing feature of valid arguments is that - if the conclusion is false, then one or more of the premises must also have been false.

An argument form is said to be valid, if and only if, when all premises are true, then the conclusion is true.

A valid argument in philosophy is characterized by the fact that if its premises are true, the conclusion must also be true. This follows a process called deductive inference. It is impossible for the premises of a valid argument to be true and the conclusion to be false

One distinguishing feature of valid arguments is that if the premises are true, then the conclusion must also be true. This follows a process called deductive inference. A premise in an argument is a statement or assumption that forms the basis for a conclusion. An argument is set to be valid when its structure or form guarantees the truth of the conclusion if the premises upon which it's built are true.

In a valid argument, therefore, the truth of its premises guarantees the truth of its conclusion-it's impossible for the premises to be true and the conclusion to be false at the same time.

If you can construct a scenario where the premises are true but the conclusion is false, then that argument is invalid. Understanding this concept is fundamental to the evaluation of arguments in logic, for it plays a critical role in ensuring coherent, reasonable and valid reasoning.

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Answer:

f(t)=9,220,000(0.888)^t

Step-by-step explanation:

We use 9,220,000 as our base because this is where the decay begins. To get the correct amount of decay, we need to subtract 11.2% from 1, or 100%. We get 88.8% so we turn this into a decimal by dividing by 100 and getting 0.888. We put this to the power of t to represent the decay because it has to be an exponent to be exponential decay, not linear decay. f(t) is not necessary, you could also use y. Hope this helps :)

Answer:

  "unchanged"

Step-by-step explanation:

Congruence transformations (translation, rotation, reflection) do not change lengths, angles, area (of 2D figures), or volume (of 3D figures). The area would remain unchanged.

Answer: There are 6045 millions world population during the period of 1950 to 2000.

Step-by-step explanation:

Since we have given that

The world population in the second half of the 20 the century by the equation:

[tex]P(t)=2560e^{0.017185t}[/tex]

We need to find the average world population during the period of 1950 to 2000.

So, there are 50 years between 1950 to 2000.

So, t = 50 years.

Therefore, the  average population would be

[tex]P(50)=2560e^{0.017185\times 50}\\\\P(50)=6045.15\\\\P(50)\approx 6045\ millions[/tex]

Hence, there are 6045 millions world population during the period of 1950 to 2000.

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus [tex]11 - 1 = 10[/tex].

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

[tex]t \approx 1.812[/tex].

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

[tex]\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}}[/tex],

where

For this confidence interval:

Hence the width of the 90% confidence interval is

[tex]\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65[/tex].

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

[tex](38.3, 53.7)[/tex].

Answer: Last option.

Step-by-step explanation:

Given the inequality [tex]|x| > 5[/tex] you need to set up two posibilities. These posibilities are the following:

- FIRST POSIBILITY :

[tex]x>5[/tex]

- SECOND POSIBILTY:

 [tex]x<-5[/tex]

Then you get that:

 [tex]x<-5\ or\ x>5 [/tex]  

Therefore, through this procedure, you get that the solution set of the inequality  [tex]|x| > 5[/tex]  is this:

{[tex]x|x<-5\ or\ x>5[/tex]}

Answer:

C. {x|x < -5 or x > 5}

Step-by-step explanation:

Answer: 30 cards

Step-by-step explanation:

To find the greatest possible number of cards in each pile such that there is the same number of cards in each​ pile we need to find the greatest common factor of 750 and 660.

Using Euclid division method , we have

[tex]750=1(660)+90\\\\660=7(90)+30\\\\90=3(30)+0[/tex]

Hence, the greatest common factor of 750 and 660 = 30

Thus, there will be 30 cards in each pile.

Answer:A

Step-by-step explanation:

Answer:

3.73%

Step-by-step explanation:

The formula is

[tex]A=P(1+\frac{r}{n})^{tn}[/tex]

Here we are given that A= 5809.81 , P=4000 , t=10 years and n = 4 (compounded quaterly)

Now we have to substitute them in the formula

[tex]5809.81=4000(1+\frac{r}{4})^{40}[/tex]

[tex]\frac{5809.81}{4000}=(1+\frac{r}{4})^{40}[/tex]

[tex] (\frac{5809.81}{4000})^{\frac{1}{40}}=1+\frac{r}{4}[/tex]

[tex] (1.45)^{\frac{1}{40}}=1+\frac{r}{4}[/tex]

[tex]1.0093 = 1+\frac{r}{4}[/tex]

Subtracting 1 on both sides

[tex]0.0093=\frac{r}{4}[/tex]

[tex]r=0.0093*4[/tex]

r=0.03732

Rate is 3.73%

Answer:

[tex]\large\boxed{5y^4+11y^2+2=(y^2+2)(5y^2+1)}[/tex]

Step-by-step explanation:

[tex]5y^4+11y^2+2=5y^4+10y^2+y^2+2=5y^2(y^2+2)+1(y^2+2)\\\\=(y^2+2)(5y^2+1)[/tex]

Answer: Hence, the value of real output produced Y = 900.

Step-by-step explanation:

Since we have given that

[tex]Y=C+I_g+G+Xn------------(1)[/tex]

and

[tex]C=50+0.9Y-----------------(2)[/tex]

And

[tex]I_g=30\\\\G=0\\\\X_n=10[/tex]

We need to find the value of Y:

From eq(1) and eq(2), we get that

[tex]Y=50+0.9Y+I_g+G+X_n\\\\Y-0.9Y=50+30+0+10\\\\0.1Y=90\\\\Y=\dfrac{90}{0.1}\\\\Y=900[/tex]

Hence, the value of real output produced Y = 900.

Answer:

The mean of the data is 26.

Step-by-step explanation:

Consider the provided data:

30, 29, 28, 30, 24, 12, 26, 33, 25, 23

Mean can be calculated as:

The average of all the data in a set.

[tex]x=\frac{\Sigma x_{i}}{n}[/tex]

Therefore,

[tex]x=\frac{30+29+28+30+24+12+26+33+25+23}{10}[/tex]

[tex]x=\frac{260}{10}[/tex]

[tex]x=26[/tex]

Thus, the mean of the data is 26.

Answer:

[tex]\boxed{\text{\$581.25; \$81.25}}[/tex]

Step-by-step explanation:

The formula for the total accrued amount is

A = P(1 + rt)

Data:

P = $500

r = 6.5 % = 0.065

t = 30 mo

Calculations:

(a) Convert months to years

t = 30 mo × (1 yr/12 mo) = 2.5 yr

(b) Calculate the accrued amount

A = 500(1 + 0.065 × 2.5)

  = 500(1 + 0.1625)

  = 500 × 1.1625

  = 581.25

[tex]\text{Dasha will have to pay back }\boxed{\textbf{\$581.25}}[/tex]

(c) Calculate the accumulated interest

[tex]\begin{array}{rcl}A & = & P + I\\581.25 & = & 500 + I\\I & = & 81.25\\\end{array}\\\text{The accumulated interest is }\boxed{\textbf{\$81.25}}[/tex]

Answer:

True

Step-by-step explanation:

6m (2m + 10)

= (6m)(2m) + (6m)(10)

= 12m² + 60m

= 4(3m² + 15m)

Because we factored out a "4" in the expression, by observation, we can see that regardless of what value of integer m is chosen, the entire expression is divisible by the "4" which has been factored out of the parentheses. Hence it is True.

What is the divisibility test of 4?

If a number's last two digits are divisible by 4, the number is a multiple of 4 and totally divisible by 4.

6m(2m+10) = 6m cannot be said to be always divisible by 4 but is divisible by 2 as 6m = 2*3m

(2m+10) cannot say if it is divisible by but can say is divisible by 2 as we can take 2 commons from both the integer and rewrite it as 2*(m+5).

Hence multiplying both the numbers we get = 2*3m*2(m+5) and now we group 2's together we get 4*3m*(m+5), as the whole number is multiplied by 4 it is at least divisible by 4 once.

Hence proved that 6m(2m+10) is divisible by 4 and the answer is true.

learn more about Divisibility here

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Answer:

Here is one idea:

You can use a regression line to predict what will happen at a time not given by your data.  You can use it to make predictions about future times.

Here is another one:

The equation for the regression line or even the graph of can tell us if the data is increasing as we move through future times or if it is decreasing as we move through future times.

Try to come up with one of your own so you can have three. :)

The number of possible arrangements is : 120

We know that in order to find the different arrangements which are possible we need to use the method of permutation.

The arrangement of a given word is calculated as the ratio of the factorial of number of letters to the product of factorial of numbers the number of times each letter is repeating.

The word is given to be:

                    ALASKA

There are a total of 6 letters in the given word

Out of which a letter "A" is  repeating  3 times.

Hence, the number of arrangements possible are:

        [tex]=\dfrac{6!}{3!}\\\\\\=\dfrac{6\times 5\times 4\times 3!}{3!}\\\\\\=6\times 5\times 4\\\\\\=120[/tex]

Answer: 1527

Step-by-step explanation:

Given: Mean : [tex]\mu = 3234\text{ grams}[/tex]

Standard deviation : [tex]\sigma=871\text{ grams}/tex]

Sample size : [tex]n=1600[/tex]

The formula to calculate the z score is given by :-

[tex]z=\dfrac{X-\mu}{\sigma}[/tex]

For X=1492

[tex]z=\dfrac{1492-3234}{871}=-2[/tex]

The p-value of z =[tex]P(z<-2)=0.0227501[/tex]

For X=4976

[tex]z=\dfrac{4976-3234}{871}=2[/tex]

The p-value of z =[tex]P(z<2)=0.9772498[/tex]

Now, the probability of the newborns weighed between 1492 grams and 4976 grams is given by :-

[tex]P(1492<X<4976)=P(X<4976)-P(X<1492)\\\\=P(z<2)-P(z<-2)\\\\=0.9772498-0.0227501\\\\=0.9544997[/tex]

Now, the number  newborns who weighed between 1492 grams and 4976 grams will be :-

[tex]1600\times0.9544997=1527.19952\approx1527[/tex]

Step-by-step explanation:

The probability he makes both shots is:

P = (0.90)^2

P = 0.81

There's a 81% probability he makes both shots.

Answer:

any length less then 5 ft

Answer:

[tex]\large\boxed{43^o}[/tex]

Step-by-step explanation:

[tex]\text{Faucets}\to12\%\\\\p\%=\dfrac{p}{100}\to12\%=\dfrac{12}{100}=0.12\\\\12\%\ \text{of}\ 360^o\to0.12\cdot360^o=43.2^o\approx43^o[/tex]

Answer: [tex]43^{\circ}[/tex]

Step-by-step explanation:

From the given pie-chart, the percentage for faucets = 12 %

We know that every circle has angle of [tex]360^{\circ}[/tex].

Now, the central angle for faucets is given by :-

[tex]\text{Central angle}=\dfrac{\text{Percent of Faucets}}{\text{100}}\times360^{\circ}\\\\\Rightarrow\ \text{Central angle}=\dfrac{12}{100}\times360^{\circ}=43.2^{\circ}\approx43^{\circ}[/tex]

Hence, the measure of the central angle for faucets [tex]\approx43^{\circ}[/tex]

If a set [tex]A[/tex] contains [tex]n[/tex] elements, then the number of subsets of this set is [tex]|\mathcal{P}(A)|=2^n[/tex].

[tex]A=\{4,5,6\}\\|A|=3[/tex]

[tex]|\mathcal{P}(A)|=2^3=8[/tex]

Answer:

The value of [tex]f^{-1}(2)=8[/tex]

Step-by-step explanation:

* Lets revise how to find the inverse function

- At first write the function as y = f(x)

- Then switch x and y

- Then solve for y

- The domain of f(x) will be the range of f^-1(x)

- The range of f(x) will be the domain of f^-1(x)

* Now lets solve the problem

- The inverse of the logarithmic function is an exponential function

∵ [tex]f(x)=log_{4}(x + 8)[/tex]

- Write the function as y = f(x)

∴ [tex]y=log_{4}(x+8)[/tex]

- Switch x and y

∴ [tex]x=log_{4}(y+8)[/tex]

- Lets solve it to find y

# Remember: [tex]log_{a}b=n=====a^{n}=b[/tex]

- Use this rule to find y

∴ [tex]4^{x}=(y + 8)[/tex]

- Subtract 8 from both sides

∴ [tex]4^{x}-8=y[/tex]

∴ [tex]f^{-1}(x)=4^{x}-8[/tex]

- Lets substitute x by 2

∴ [tex]f^{-1}(2)=4^{2}-8[/tex]

- The value of 4² = 16

∴ [tex]f^{-1}(x)=16-8=8[/tex]

* The value of [tex]f^{-1}(2)=8[/tex]

To find the inverse of the function f(x) = log base 4 of (x+8) when evaluated at 2, we solve the equation log base 4 of (x+8) = 2 to find x. This gives us x = 8, so f^-1(2) = 8.

To find the value of f-1(2) for the function f(x) = log4(x+8), we first need to understand that f-1(x) denotes the inverse function of f(x). This means that if f(a) = b, then f-1(b) = a. To solve for f-1(2), we set f(x) equal to 2 and solve for x:

log4(x+8) = 2

42 = x+8 (using the property that if logb(a) = c then bc = a)

16 = x+8

x = 16 - 8

x = 8

Therefore, f-1(2) = 8.

a. All points on the board are equally likely to be hit with a probability of 1/(area of board), or

[tex]f_{X,Y}(x,y)=\begin{cases}\dfrac1{4\pi}&\text{for }x^2+y^2\le4\\\\0&\text{otherwise}\end{cases}[/tex]

b. To find the marginal distribution of [tex]X[/tex], integrate the joint distribution with respect to [tex]y[/tex], and vice versa. We can take advantage of symmetry here to compute the integral:

[tex]\displaystyle\int_y f_{X,Y}(x,y)\,\mathrm dy=2\int_0^{\sqrt{4-x^2}}\frac{\mathrm dy}{4\pi}=\frac{\sqrt{4-x^2}}{2\pi}[/tex]

[tex]f_X(x)=\begin{cases}\dfrac{\sqrt{4-x^2}}{2\pi}&\text{for }-2\le x\le2\\\\0&\text{otherwise}\end{cases}[/tex]

and by the same computation you would find that

[tex]f_Y(y)=\begin{cases}\dfrac{\sqrt{4-y^2}}{2\pi}&\text{for }-2\le y\le2\\\\0&\text{otherwise}\end{cases}[/tex]

c. We get the conditional distributions by dividing the joint distributions by the respective marginal distributions:

[tex]f_{X\mid Y=y}(x)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}[/tex]

[tex]f_{X\mid Y=y}(x)=\begin{cases}\dfrac1{2\sqrt{4-y^2}}&\text{for }-2\le y\le2\text{ and }x^2\le4-y^2\\\\0&\text{otherwise}\end{cases}[/tex]

and similarly,

[tex]f_{Y\mid X=x}(y)=\begin{cases}\dfrac1{2\sqrt{4-x^2}}&\text{for }-2\le x\le2\text{ and }y^2\le4-x^2\\\\0&\text{otherwise}\end{cases}[/tex]

d. You can compute this probability by integrating the joint distribution over a part of the circle (call it "B" for bullseye):

[tex]\displaystyle\iint_Bf_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^{0.25}\frac r{4\pi}\,\mathrm dr\,\mathrm d\theta=\frac1{64}[/tex]

(using polar coordinates) The easier method would be to compute the area of a circle with radius 0.25 instead, then divide that by the total area of the dartboard.

[tex]\dfrac{\pi\left(\frac14\right)^2}{\pi\cdot2^2}=\dfrac1{64}[/tex]

e. The event that [tex]X>1[/tex] is complementary to the event that [tex]X\le1[/tex], so

[tex]P(X>1)=1-P(X\le1)=1-F_X(1)[/tex]

where [tex]F_X(x)[/tex] is the marginal CDF for [tex]X[/tex]. We can compute this by integrate the marginal PDF for [tex]X[/tex]:

[tex]F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<-2\\\\\dfrac12+\dfrac1\pi\sin^{-1}\dfrac x2+\dfrac{x\sqrt{4-x^2}}{4\pi}&\text{for }-2\le x<2\\\\1&\text{for }x\ge2\end{cases}[/tex]

Then

[tex]P(X>1)=1-F_X(1)=\dfrac13-\dfrac{\sqrt3}{4\pi}\approx0.1955[/tex]

f. We found that either random variable conditioned on the other is a uniform distribution. In particular,

[tex]f_{Y\mid X=1}(y)=\begin{cases}\dfrac1{2\sqrt3}&\text{for }y^2\le3\\\\0&\text{otherwise}\end{cases}[/tex]

Then

[tex]P(Y>0.3\mid X=1)=1-P(Y\le0.3\mid X=1)=1-F_{Y\mid X=1}(0.3)[/tex]

where [tex]F_{Y\mid X=x}(y)[/tex] is the CDF of [tex]Y[/tex] conditioned on [tex]X=x[/tex]. This is easy to compute:

[tex]F_{Y\mid X=1}(y)=\displaystyle\int_{-\infty}^yf_{Y\mid X=1}(t)\,\mathrm dt=\begin{cases}0&\text{for }y<-\sqrt3\\\\\dfrac{y+\sqrt3}{2\sqrt3}&\text{for }-\sqrt3\le y<\sqrt3\\\\1&\text{for }y\ge\sqrt3\end{cases}[/tex]

and we end up with

[tex]P(Y>0.3\mid X=1)=\dfrac{10-\sqrt3}{20}\approx0.4134[/tex]

Final answer:

The joint PDF for a uniform distribution on a circular dart board is constant within the dart board and zero outside. Marginal and conditional PDFs are derived from the joint PDF. To compute probabilities such as the bulls eye hit or X > 1, integrate the corresponding PDF over the relevant range.

Explanation:

Finding the Joint and Marginal PDFs, and Conditional Probabilities

The joint probability density function (PDF) of X and Y for a uniform distribution over a circular dart board with radius 2 inches is constant within the circle and zero outside. First, we calculate the area of the circle, A = πr² = π(2)² = 4π square inches. The joint PDF f(x, y) will be 1/A for all points inside the dart board, and 0 otherwise.

The marginal PDFs are derived by integrating the joint PDF over the other variable. For instance, fX(x) is found by integrating f(x, y) over y, and fY(y) is found by integrating f(x, y) over x.

The conditional PDFs fX|Y and fY|X are derived from the joint PDF divided by the marginal PDF of the conditioned variable.

To find the probability of hitting the bulls eye, a circle of radius 0.25 inches, you'd integrate the joint PDF over the area of the bulls eye or simply calculate the area ratio of the bulls eye to that of the entire dart board.

The probability that X > 1 is found by integrating fX(x) from 1 to 2. If you know that X = 1, the conditional PDF of Y is fY|X(y|X=1). The probability that Y > 0.3 given X = 1 is calculated by integrating this conditional PDF from 0.3 to the upper limit set by the circle's boundary.