A Bernoulli random variable X has unknown success probability p. Using 100 independent samples of X, find a confidence interval estimate of p with confidence coefficient 0.99. If ????????100 = 0.06, what is our interval estimate

Set

[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta[/tex]

The region [tex]R[/tex] is given in polar coordinates by the set

[tex]R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}[/tex]

So we have

[tex]\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac\pi4(\cos4-\cos9)}[/tex]

The Cartesian coordinates are converted into polar coordinates so that the integral sin(x2 + y2) dA R becomes the integral sin(r2) r dr dθ. But this specific integral can't be solved analytically, yet using polar coordinates can simplify other integration issues related to circular regions or distances from the origin.

To evaluate the given integral using polar coordinates, one must firstly translate the Cartesian coordinates (x,y) into polar coordinates (r,θ), so that x is replaced with rcosθ and y with rsinθ. Consequently, the integral sin(x2 + y2) dA R becomes the integral sin(r2) r dr dθ, with r varying from 2 to 3, and θ from 0 to π/2 (since we are only dealing with the first quadrant).

However, this integral becomes very complex and is not feasible to solve analytically. You would need to use a numeric method to get an approximate answer. In the context of other problems, switching to polar coordinates can simplify the integration, especially when dealing with circular regions or equations related to distances from the origin.

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Answer:

Step-by-step explanation:

See if the attachment below helps you with this.

Answer:

The solution of the given equation is [tex]\sqrt[3]{4}-\sqrt[3]{2}[/tex].

Step-by-step explanation:

According to the cardano's method, the solution of the equation is x=u-v. If the equation is

[tex]x^3+px=q[/tex]

Where [tex]u^3-v^3=q[/tex]

[tex]3uv=p[/tex]

The given equation is

[tex]x^3+6x=2[/tex]

Here p=6 and q=2.

[tex]u^3-v^3=2[/tex]                 .... (1)

[tex]3uv=6[/tex]

[tex]uv=2[/tex]

Taking cube both the sides.

[tex]u^3v^3=8[/tex]

Multiply both sides by 4.

[tex]4u^3v^3=32[/tex]             .... (2)

Taking square both the sides of equation (1).

[tex](u^3-v^3)^2=2^2[/tex]

[tex](u^3)^2-2u^3v^3+(v^3)^2=4[/tex]       .... (3)

Add equation (2) and (3).

[tex](u^3)^2-2u^3v^3+(v^3)^2+4u^3v^3=4+32[/tex]

[tex](u^3+v^3)^2=36[/tex]

Taking square root both the sides.

[tex]u^3+v^3=6[/tex]             .... (4)

On adding equation (1) and (4), we get

[tex]2u^3=8[/tex]

[tex]u^3=4[/tex]

[tex]u=\sqrt[3]{4}[/tex]

On subtracting equation (1) and (4), we get

[tex]-2v^3=-4[/tex]

[tex]v^3=2[/tex]

[tex]v=\sqrt[3]{2}[/tex]

The solution of the equation is

[tex]x=u-v=\sqrt[3]{4}-\sqrt[3]{2}[/tex]

Therefore the solution of the given equation is [tex]\sqrt[3]{4}-\sqrt[3]{2}[/tex].

Final answer:

Solving x³+6x=2 using Cardano's method involves rewriting the equation to match the standard form of a depressed cubic equation, calculating the required constants, and finally, applying these constants to find the roots.

First, let's rewrite the equation x³+6x-2 = 0 as x³+6x = 2 to match the standard form of a depressed cubic equation which is x³ +px = q. Here, p = 6 and q = 2.

Next, we calculate the value t = sqrt[(q/2)² + (p/3)³]. So, t = sqrt[(1)² + (2)³] = sqrt[1 + 8] = 3.

Using these values, we can now calculate the roots. We know the roots are given by the formulaes u-v where u = cubicroot(q/2 + t) and v = cubicroot(q/2 - t). So, u = cubicroot(1 + 3) = 2, and v = cubicroot(1 - 3) = - root(2).

Therefore, the roots of the given polynomial equation are x = u - v = 2 - (- root(2)) = 2 + root(2).

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We have to show that

[tex]\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}[/tex]

for [tex]\frac{\partial ^{2}u}{\partial t^{2}}[/tex] we have

[tex]\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}[/tex]

[tex]\frac{\partial ^{2}u}{\partial t^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial t^{2}}[/tex]

[tex]=\frac{\partial }{\partial t}[\frac{\partial[f(x-at)+g(x+at)] }{\partial t}][/tex]

[tex]\frac{\partial }{\partial t}[-a\cdot f'(x-at)+a\cdot g'(x+at)][/tex]

[tex]=a^{2}f''(x-at)+a^{2}g''(x+at)[/tex]

[tex]=a^{2}[f''(x-at)+g''(x+at)].............(i)[/tex]

similarly,

[tex]\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial x^{2}}[/tex]

[tex]=\frac{\partial }{\partial x}[\frac{\partial[f(x-at)+g(x+at)] }{\partial x}][/tex]

[tex]=\frac{\partial }{\partial x}[f'(x-at)+g'(x+at)][/tex]

[tex]=f''(x-at)+g''(x+at).......(ii)[/tex]  

Comparing i and ii we get  

[tex]a^{2}\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}u}{\partial t^{2}}[/tex]

Hence proved

There are 16 numbers.

The median is 92.5 ( find the middle two values and divide by 2).

Minumum is 81

Maximum is 109

First quartile is 88.25 (Find median of the lower half of numbers).

Third quartile is 97.75 (Find median of the upper half of numbers.)

The interquartile range is 9.5 ( Difference between the first and third quartile).

Plotting that data in a box plot, the correct one looks like #1

Find the median.

92 and 93 are both middle numbers so add and divide by two.

92 + 93 = 185

185 / 2 = 92.5

Minimum (smallest number): 81

Maximum (largest number): 109

Find the median of the lower values behind the median.

88.25

Find the mean of the higher values ahead of the median.

97.75

Subtract to find the interquartile range.

97.75 - 88.25 = 9.5

The only option with these characteristics is Option A.

Best of Luck!

Answer: They meet after 1 hour 42 minutes.

Step-by-step explanation:

Since we have given that

Dana leaves Las Vegas for LA at 2 p.m. driving at 55 mph.

Let the time taken by Dana be 't'.

Distance traveled by Dana would be 55t.

At 4 p.m. Lance leaves LA for Las Vegas driving at 45 mph along the same route.

It means after 2 hours Lance leave for LA.

So, time taken by Lance be 't-2'.

Distance traveled by Lance would be 45(t-2)

Total distance  = 260 miles

According to question, it becomes,

[tex]55t+45(t-2)=260\\\\55t+45t-90=260\\\\100t=260-90\\\\100t=170\\\\t=1.7\ hours=1\dfrac{7}{10}=1\ hour\ and\ \dfrac{7\times 60}{10}\ minutes=1\ hour\ 42\ minutes[/tex]

Hence, they meet after 1 hour 42 minutes.

Answer: i dont now

Step-by-step explanation:

u have to add them togther i guess

Answer:

$5.80 Option B.

Step-by-step explanation:

It is given that a 20% tip on a meal that costs $29.17.

The cost of the meal = $29.17

Tip on a meal = 20%

Therefore, 20% of $29.17

= [tex]\frac{20}{100}[/tex] × 29.17

= 0.20 × 29.17

= 5.834

= $5.80

The correct estimate would be $5.80 Option B.

Answer: There is 162 ml of first brand and 108 ml of second brand.

Step-by-step explanation:

Since we have given that

Percentage of vinegar that the first brand contains = 7%

Percentage of vinegar that the second brand contains = 12%

Percentage of vinegar in mixture = 9%

Total amount of dressing = 270 ml

We will use "Mixture and Allegation":

First brand                  Second brand

     7%                                12%

                       9%

--------------------------------------------------------

12%-9%             :                9%-7%

 3%                   :                    2%

So, ratio of first brand to second brand in a mixture is 3:2.

So, Amount of first brand she should use is given by

[tex]\dfrac{3}{5}\times 270\\\\=162\ ml[/tex]

Amount of second brand she should use is given by

[tex]\dfrac{2}{5}\times 270\\\\=108\ ml[/tex]

Hence, there is 162 ml of first brand and 108 ml of second brand.

Answer:

7/32

Step-by-step explanation:

Answer:

The correct option is d.

Step-by-step explanation:

The approximate population P, in thousands, for a species of frogs in a particular rain forest, x years after 1999 is given by the formula

[tex]P=0.672x^2-0.046x+3[/tex]

We need to find the year it which the population reach 182 frogs.

Substitute P=182 in the given formula.

[tex]182=0.672x^2-0.046x+3[/tex]

Subtract 182 from both the sides.

[tex]0=0.672x^2-0.046x+3-182[/tex]

[tex]0=0.672x^2-0.046x-179[/tex]

Multiply both sides by 1000 to remove decimals.

[tex]0=672x^2-46x-179000[/tex]

Quadratic formula:

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

Substitute a=672, b=-46 and c=-179000 in the quadratic formula.

[tex]x=\frac{-\left(-46\right)\pm\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}[/tex]

On simplification we get

[tex]x=\frac{-\left(-46\right)+\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx 16.355[/tex]

[tex]x=\frac{-\left(-46\right)-\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx -16.287[/tex]

The value of x can not be negative because x is number of years after 1999.

x=16.35 in means is 17th year after 1999 the population reach 182 frogs.

[tex]1999+17=2016[/tex]

The population reach 182 frogs in 2016. Therefore the correct option is d.

Answer:

B. 78 and 82.

Step-by-step explanation:

We have been given that the average test score of the class was an 80 and the standard deviation was 2. We are asked to find two values between which 68% of class will score.

We know that in a normal distribution approximately 68% of the data falls within one standard deviation of the mean.

So 68% scores will lie within one standard deviation below and above mean that is:

[tex](\mu-\sigma,\mu+\sigma)[/tex]

Upon substituting our given values, we will get:

[tex](80-2,80+2)[/tex]

[tex](78,82)[/tex]

Therefore, about 68% of the class would score between 78 and 81 and option B is the correct choice.

Answer with explanation:

The System of equations which we have to solve by Gauss Jordan Method:

  [tex]1.\rightarrow 2x_{1}-6x_{2}-2x_{3}=14, 2.\rightarrow 3x_{1}+4x_{2}-7x_{3}= 16, 3.\rightarrow 3x_{1}-6x_{2}+9x_{3}=21[/tex]

Writing it in the form of Augmented Matrix=3 Rows and 4 Columns:

  [tex]\left[\begin{array}{cccc}2&-6&-2&14\\3&4&-7&16\\3&-6&9&21\end{array}\right]\\\\R_{1}=\frac{R_{1}}{2},R_{3}=\frac{R_{3}}{3}\\\\ \left[\begin{array}{cccc}1&-3&-1&7\\3&4&-7&16\\1&-2&3&7\end{array}\right]\\\\R_{3}\rightarrow R_{3}-R_{1}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\3&4&-7&16\\0&1&4&0\end{array}\right]\\\\R_{2}\rightarrow R_{2}-3R_{1}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\0&13&-4&-5\\0&1&4&0\end{array}\right][/tex]

 [tex]R_{3}\rightarrow R_{2}+R_{3}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\0&13&-4&-5\\0&14&0&-5\end{array}\right]\\\\\rightarrow14 x_{2}= -5\\\\x_{2}=\frac{-5}{14}\\\\\rightarrow 13 x_{2}-4x_{3}=-5\\\\ \frac{-65}{14}-4 x_{3}=-5\\\\-4x_{3}=-5+\frac{65}{14}\\\\x_{3}=\frac{5}{56}\\\\x_{1}-3x_{2}-x_{3}=7\\\\x_{1}+\frac{15}{14}-\frac{5}{56}=7\\\\x_{1}+\frac{55}{56}=7\\\\x_{1}=7-\frac{55}{56}\\\\x_{1}=\frac{337}{56}[/tex]

Solution set

  [tex]=(\frac{337}{56},\frac{-5}{14},\frac{5}{56})[/tex]

Answer: 0.0222

Step-by-step explanation:

Given : The distribution of trout weights is normally distributed with

Mean : [tex]\mu=402.7\text{ grams}[/tex]

Standard deviation : [tex]\sigma=8.8\text{ grams}[/tex]

Sample size : [tex]n=40[/tex]

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

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

Let x be the weight of randomly selected trout.

Then for x = 405.5  , we have

[tex]z=\dfrac{405.5 -402.7}{\dfrac{8.8}{\sqrt{40}}}\approx2.01[/tex]

The p-value : [tex]P(405.5<x)=P(2.01<z)[/tex]

[tex]1-P(2.01)=1-0.9777844=0.0222156\approx0.0222[/tex]

Thus,the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams= 0.0222.

The probability that the mean weight for a sample of 40 trout exceeds 405.5 grams is  0.0222.

The distribution of trout weights is normally distributed with

We have given that

Mean=402.7 grams

Standard deviation =8.8  grams  

Sample size (n)=40

We have to calculate

The probability that the mean weight for a sample of 40 trout exceeds 405.5 grams

Te formula of Z score is given by,

[tex]z=\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}[/tex]

n= the ample size

x=mean

sigma=standard deviation

So by using the formula we have,

Let x is  the weight of randomly selected trout.

Then for x = 405.5  

[tex]z=\frac{405.5-\402.7}{\frac{\8.8 }{\sqrt{40}}}\\\\\z=2.01[/tex]

we have

The p-value :(405.5<x)

(1-2.01)=1-0.97778

           =0.02221

           =0.0222

Therefore,the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams= 0.0222.

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Answer:

  linear function growth rate: 8.5

Step-by-step explanation:

The growth rate of the linear function is the coefficient of t: 8.5. (It is a constant.)

__

The growth rate of g(t) is its derivative: g'(t) = (1/8)(160e^(t/8)) = 20e^(t/8). Then the relative growth rate is ...

  g'(t)/g(t) = (20e^(t/8))/(160e^(t/8)) = 20/160 = 1/8

It is a constant.

Answer:a-396

b-420

Step-by-step explanation:

[tex]\alpha[/tex] =0.1

Margin of Error=0.04

Level of significance is z[tex]\left ( 0.1\right )=1.64[/tex]

Previous estimate[tex]\left ( p\right ) =0.38[/tex]

sample size is given by:

n=[tex]\left (\frac{Z_{\frac{\alpha }{2}}}{E}\right )p\left ( 1-p\right )[/tex]

n=[tex]\frac{1.64}{0.04}^{2}0.38\left ( 1-0.38\right )=396.0436\approx 396[/tex]

[tex]\left ( b\right )[/tex]Does not use prior estimate

Assume

[tex]\alpha [/tex]=0.1

Margin of Error=0.04

Level of significance is z[tex]\left ( 0.1\right )=1.64[/tex]

Population proportion[tex]\left ( p\right )[/tex]=0.5

n=[tex]\left (\frac{Z_{\frac{\alpha }{2}}}{E}\right )p\left ( 1-p\right )[/tex]

n=[tex]\frac{1.64}{0.04}^{2}0.5\left ( 1-0.5\right )[/tex]

n=420.25[tex]\approx 420[/tex]

Answer: Option(c) Richard Lester is correct.

Step-by-step explanation:

Both the films were directed by Richard Lester.

A Hard Day's night was a scripted comic farce and its main focus on  Beatlemania and the band's hectic touring lifestyle. It is a black and white movie.

Help! film also directed by Richard Lester. And this film was shot in various exotic locations. Help! was the first Beatles film that is filmed in colour.  

Answer: 0.7257

Step-by-step explanation:

Given : The weights of steers in a herd are distributed normally.

[tex]\mu= 1100\text{ lbs }[/tex]

Standard deviation : [tex]\sigma=300 \text{ lbs }[/tex]

Let x be the weight of the randomly selected steer .

Z-score : [tex]\dfrac{x-\mu}{\sigma}[/tex]

[tex]z=\dfrac{920-1100}{300}=-0.6[/tex]

The the probability that the weight of a randomly selected steer is greater than 920 lbs using standardized normal distribution table  :

[tex]P(x>920)=P(z>-0.6)=1-P(z<-0.6)\\\\=1-0.2742531=0.7257469\approx0.7257[/tex]    

Hence, the probability that the weight of a randomly selected steer is greater than 920lbs =0.7257

Answer:

Probability: 0.1825 or 18.25%  .......  Unlikely

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Probability problems.

To start of we need to calculate the total amount of respondents that took the survey. We do this by adding all the answers together.

169 + 12,527 + 2,834 = 15,530 total people

Now that we know the total amount of people we can calculate the probability of each response by dividing the amount of people that had that response by the total amount of people that took the survey.

Never Used: [tex]\frac{169}{15,530} = 0.0109 = 1.09%[/tex]

Sometimes Used: [tex]\frac{12,527}{15,530} = 0.8066 = 80.66%[/tex]

Frequently used: [tex]\frac{2834}{15,530} = 0.1825 = 18.25%[/tex]

So we can see that the probability of a randomly selected person using a credit card frequently is 0.1825 or 18.25%

Unlikely

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Final answer:

The probability that a randomly selected person uses a credit card frequently is calculated as 2834 divided by the total of 15530, resulting in 0.1825 (rounded to four decimal places). Since this is less than 20%, it is considered 'Unlikely' for someone to frequently use a credit card.

Explanation:

To calculate the probability that a randomly selected person uses a credit card frequently, we need to use the basic probability formula, which is the number of favorable outcomes divided by the total number of outcomes. In this case, the number of people who use a credit card frequently is the favorable outcome, and the total number of respondents is the sum of all categories of credit card usage.

Number of people who use a credit card frequently: 2834

Total number of respondents: 169 (never) + 12527 (sometimes) + 2834 (frequently) = 15530

Probability of frequent use: 2834 / 15530 = 0.1825 (rounded to 4 decimal places)

Now, let's interpret the result. A probability of 0.1825, when rounded, is about 18.25%. This number is less than 20%, which is generally considered the benchmark for something to be considered "likely". Therefore, it is Unlikely for someone to use a credit card frequently.

Approximately 2.28% of pediatricians work more than 72 hours per week according to the table.

To find the percentage of pediatricians who work more than 72 hours per week, we need to calculate the z-score for this value and then use a standard normal distribution table to find the corresponding percentage.

Calculate the z-score using the formula:

[tex]z = (x - u) / \alpha[/tex]

where x is the value (72 hours), u is the mean (48 hours), and a is the standard deviation (12 hours).

Substitute the values into the formula: z = (72 - 48) / 12 = 2.

Using a standard normal distribution table, find the percentage of values that are greater than 2.

Based on the table, approximately 2.28% of pediatricians work more than 72 hours per week.

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Answer:

[tex]\large\boxed{B=A+2I}[/tex]

Step-by-step explanation:

It's possible if dimensions of a matrix A and matrix B are n × n

[tex]AB=A^2+2A\qquad\text{multiply both sides on the left by}\ A^{-1}\\\\A^{-1}AB=A^{-1}A^2+A^{-1}(2A)\qquad\text{we know}\ A^{-1}A=I\\\\IB=A^{-1}A\cdot A+2A^{-1}A\\\\IB=IA+2I\qquad\text{we know}\ IA=A\\\\B=A+2I[/tex]

Matrices is an array of numbers, usually 2 dimensional, but can be single dimensional too.

A matrix B such that [tex]AB = A^2 + 2A[/tex] is given as

[tex]B = A + 2I = \left[\begin{array}{cc}4&0\\0&4\end{array}\right][/tex]

(Assuming A is left invertible and  [tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex])

Suppose that there is an equation

[tex]AB = AC[/tex]

We cannot always say that [tex]B = C[/tex]

If we assume that A is left invertible, then only we can surely say that we have got [tex]B = C[/tex]

Similarly, for [tex]BA = CA[/tex] to imply  [tex]B = C[/tex], we need A to be right invertible.

Assuming that we have A as a left invertible matrix, say

[tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex]

and [tex]L_A[/tex] be its left inverse, then [tex]L_A A = I[/tex] ([tex]I[/tex] is identity matrix)

Then,

[tex]AB = A^2 + 2A\\A(B) = A(A + 2I_2)\\\\\Multiplying L_{A}\text{ on left side of both terms,}\\\\L_{A} AB = L_{A}A(A + 2I_2)\\B = A + 2I_2\\\\B = \left[\begin{array}{cc}2&0\\0&2\end{array}\right] + \left[\begin{array}{cc}2&0\\0&2\end{array}\right] = \left[\begin{array}{cc}4&0\\0&4\end{array}\right] = 4I_2\\\\B = 4I_2[/tex]

Thus, i

A matrix B such that [tex]AB = A^2 + 2A[/tex] is given as

[tex]B = \left[\begin{array}{cc}4&0\\0&4\end{array}\right][/tex]

(Assuming A is left invertible and  [tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex])

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Answer:

The required function is [tex]p\left(x\right)=1.325-0.675x+0.375x^2[/tex].

Step-by-step explanation:

The given data points are (0, 1), (1, 2), (2, 1/2) and (3, 3).

Let the quadratic function is defined as

[tex]p(x)=a_0+a_1x+a_2x^2[/tex]              .... (1)

Using graphing calculator, we get

[tex]a_0=1.325[/tex]

[tex]a_1=-0.675[/tex]

[tex]a_2=0.375[/tex]

Substitute [tex]a_0=1.325[/tex], [tex]a_1=-0.675[/tex] and [tex]a_2=0.375[/tex] in function (1), to find the quadratic function.

[tex]p\left(x\right)=1.325-0.675x+0.375x^2[/tex]

Therefore the required function is [tex]p\left(x\right)=1.325-0.675x+0.375x^2[/tex].

The graph of data points and quadratic function is shown below.

Answer:

  (a)  20% to Longmont; 80% to Denver

  (b)  20% to Longmont; 80% to Denver

Step-by-step explanation:

(a) The bus to Longmont is the first bus to arrive, only between 8:31 and 8:37, and again between 9:01 and 9:07. That is, for a total of 12 minutes every hour, the Longmont bus is the first to arrive. The probability of going to Longmont is 12/60 = 1/5 = 20%.

__

(b) Same as for (a). As long as passenger arrival times are uniform within an hour, the probability is the same.

Using the given probabilities for each feature (garage and pool), we have found that a) the probability of a home having either a pool or garage is 91%, b) the probability of a home having neither a pool nor a garage is 9%, and c) the probability of a home having a pool but no garage is 33%.

The question is asking about the probability of certain features in homes for sale, namely garages and swimming pools. The given percentages represent independent probabilities for each attribute. Let's denote garage as 'G' and pool as 'P'. Then the probabilities given are P(G)=0.58, P(P)=0.39, and P(G and P)=0.06.

a) The probability a home has a pool or a garage: This is determined using the formula for the union of two events: P(G U P) = P(G) + P(P) - P(G and P) = 0.58+0.39-0.06 = 0.91 or 91% of homes for sale.

b) The probability a home has neither a pool nor a garage: This is the complement of the event in part a. So, P(Neither G nor P) = 1 - P(G U P) = 1 - 0.91 = 0.09 or 9% of homes for sale.

c) The probability a home has a pool but not a garage: This is determined using the formula for the difference of two events: P(P - G) = P(P) - P(G and P) = 0.39 - 0.06 = 0.33 or 33% of homes for sale.

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To construct a 95% confidence interval with a margin of error of 0.052, a sample size of 300 is needed.

To determine the sample size needed for the 95% confidence interval with a margin of error of 0.052, we can use the formula:

n = (Z^2 * p * (1 - p)) / (E^2)

where n is the sample size, Z is the Z-score corresponding to the desired confidence level (in this case, 1.96), p is the estimated proportion (0.33), and E is the margin of error (0.052).

Substituting the given values into the formula:

n = (1.96^2 * 0.33 * (1 - 0.33)) / (0.052^2)

Simplifying the equation:

n = 299.5554

Rounding up to the nearest whole number, the sample size needed is 300.

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To construct a 95% confidence interval for the proportion of adults who believe economic conditions are getting better with a margin of error of 0.052, the required sample size is approximately 577 participants.

To calculate the sample size needed to construct a 95% confidence interval for the proportion of adults who believe that economic conditions are getting better, with a Gallup poll estimate of 0.33 and a desired margin of error of 0.052, the formula for determining the sample size (n) is:

n = (Z^2*p*(1-p))/E^2,

where Z is the Z-score corresponding to the 95% confidence level, p is the estimated proportion (0.33) and E is the margin of error (0.052). The Z-score for a 95% confidence level is 1.96. Plugging in the values gives:

n = (1.96^2*0.33*(1-0.33))/(0.052^2),

Solving this, we find the sample size required:

n= 576.7.

Since we cannot have a fraction of a person, we round up to the next whole number. Therefore, the sample size needed is 577 participants.

I'll abbreviate the definite integral with the notation,

[tex]I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx[/tex]

We're given

Recall that the definite integral is additive on the interval [tex][a,b][/tex], meaning for some [tex]c\in[a,b][/tex] we have

[tex]I(f,a,b)=I(f,a,c)+I(f,c,b)[/tex]

The definite integral is also linear in the sense that

[tex]I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)[/tex]

for some constant scalars [tex]k,\ell[/tex].

Also, if [tex]a\ge b[/tex], then

[tex]I(f,a,b)=-I(f,b,a)[/tex]

a. [tex]I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}[/tex]

b. [tex]I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}[/tex]

c. [tex]I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}[/tex]

d. [tex]I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}[/tex]

e. [tex]I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}[/tex]

f. [tex]I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}[/tex]

In this integral calculus problem, we leverage properties of definite integrals to compute the values of various expressions. Key steps usually involve substituting given integral values and multiplying by constant factors when required

To solve the problem, we first need to consider the properties of integral calculus, specifically those of definite integrals. A fundamental rule that is applicable here is that the product of a constant and an integral is the constant times the value of the integral.

So for problem a, integral^5_1 2f(x) dx = 2* integral^5_1 f(x) dx = 2 * 5 = 10.

Similarly, for problem b, integral^8_1 (f(x) - g (x)) dx = integral^8_1 f(x) dx - integral^8_1 g(x) dx = 9 - 5 = 4.

Following through similar steps of substitutions, we obtain the following solutions:

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For this case we have the following functions:

[tex]F (x) = 3x ^ 2 +1\\G (x) = 2x-3\\H (x) = x[/tex]

We have to:

[tex]G (x) * - 1[/tex] is given by:

[tex](2x-3) * - 1 = -2x +3[/tex]

Thus, the correct option is the option is A.

On the other hand,

[tex]F (x) +G (x) = 3x ^ 2 +1 +(2x - 3) = 3x ^ 2+ 1+ 2x-3 = 3x ^ 2 +2x-2[/tex]

Thus, the correct option is the option is A.

We also have:

[tex]F (-2) = 3 (-2) ^ 2+ 1 = 3 (4)+ 1 = 12+ 1 = 13[/tex]

Thus, the correct option is the option is B.

Last we have:

[tex]F (3)+G (4) -2H (5) = (3 (3) ^ 2+ 1)+ (2 (4) -3) -2 (5) = (3 (9)+ 1) - (8-3) 10 = 28+ 5-10 = 33[/tex]

Thus, the correct option is the option is C.

ANswer:

Option A, A, B, C

Answer:

G-1(x)= -2x+3

F(x)+G(x)=3x^2+2x-2

F(-2)=13

F(3)+G(4)-2H(5)=33

A,A,B,C are the answers to the equations

Answer:

Step-by-step explanation:

Assuming A as x axis and B as y axis the equations are

[tex]2x+0.5y\geq 90\quad \left(1\right)\\0.75x+5y\geq 200\quad \left(2\right)\\0.75x+1.5y\leq 150\quad \left(3\right)\\x,y\geq 0\quad \left(4\right)[/tex]

Solving equations (2) and (3) we get

x=171.429 y=14.286

Solving equations (1) and (3) we get

x=22.857 y=88.571

Solving equations (1) and (2) we get

x=36.364 y=34.545

The area enclosing the above three points is the feasible region.

Final answer:

The test statistic for the proportion of yellow pea pods being 25% is calculated using the sample proportion, hypothesized proportion, and sample size. Using the given data, the test statistic (Z-score) comes out to approximately -1.412.

Explanation:

To find the value of the test statistic for the claim that the proportion of peas with yellow pods is equal to 0.25 (25%), we use the sample statistics provided from the experiment. You mentioned that there were 590 peas in total, with 139 having yellow pods. First, we check if the conditions for the binomial distribution are met, which in this case they are as we are dealing with two outcomes (yellow pods and not yellow pods), a fixed number of trials (590 peas), and each pea is independent of the others.

The test statistic for a proportion is calculated using the formula:

Z = (p' - p) / (sqrt(p(1 - p) / n))

Where:

Now, we calculate the test statistic:

Z = (0.2356 - 0.25) / (sqrt(0.25 × (1 - 0.25) / 590))

Z ≈ -1.412

This Z-score tells us how many standard deviations the observed sample proportion (0.2356) is from the hypothesized proportion (0.25).

Answer:

Mean = 7.38

SD = 0.148

b. 95%

Step-by-step explanation:

Given data is:

7.1 7.5 7.6 7.4 7.3 7.3 7.3 7.5

Mean:

Mean = Sum/No. of values

= (7.1+7.5+7.6+7.4+7.3+7.3+7.3+7.5)/8

=59/8

=7.38

Standard Deviation:

x                  x-x'           (x-x')^2

7.1               -0.28           0.0784

7.5              0.12            0.0144

7.6              0.22           0.0484

7.4              0.02           0.0004

7.3             -0.08           0.0064

7.3             -0.08            0.0064

7.3              -0.08            0.0064

7.5              0.12              0.0144

                      Total :     0.1752

Variance = Sum of squares/No of items

= 0.1752/8 = 0.0219

SD =√0.0219 = 0.148

b. What percentage of the data is within 2 standard deviations of the​ mean?

95% of data is within two standard deviations of mean in a standard normal distribution ..

The mean of the pH test results is 7.375, and the standard deviation is approximately 0.086.

To find the mean and standard deviation of the pH test results, we can use the following formulas:

Mean: Add up all the pH values and divide by the total number of samples (in this case, 8). So, (7.1 + 7.5 + 7.6 + 7.4 + 7.3 + 7.3 + 7.3 + 7.5) / 8 = 7.375.

Standard Deviation: Calculate the difference between each pH value and the mean, square each difference, calculate the mean of those squared differences, and then take the square root. Let's break it down into steps: Subtract the mean from each pH value: (7.1 - 7.375), (7.5 - 7.375), (7.6 - 7.375), (7.4 - 7.375), (7.3 - 7.375), (7.3 - 7.375), (7.3 - 7.375), (7.5 - 7.375). Square each difference: (0.0425)^2, (0.125)^2, (0.225)^2, (0.025)^2, (-0.075)^2, (-0.075)^2, (-0.075)^2, (0.125)^2. Calculate the mean of the squared differences: (0.0018 + 0.0156 + 0.0506 + 0.000625 + 0.005625 + 0.005625 + 0.005625 + 0.0156) / 8 = 0.0074. Take the square root of the mean: √0.0074 ≈ 0.086.

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