Include ALL the steps (statements/reasons) by writing a paragraph proof.

Answer:

[tex]\frac{7}{4}[/tex]×x

Step-by-step explanation:

given ratio of yellow to red flowers is 4:7.

let number of yellow flowers in Lindsay’s vase  is x

let number of red flowers in Lindsay’s vase  is y

[tex]\frac{x}{y}[/tex]=[tex]\frac{4}{7}[/tex]

number of  red flowers such that x is known is y= [tex]\frac{7}{4}[/tex]×x

Answer:

(-0.0861, 0.4023)

Step-by-step explanation:

We have large sample sizes [tex]n_{x} = 56[/tex] and [tex]n_{y} = 41[/tex]. A [tex]100(1-\alpha)[/tex]% confidence interval for the difference [tex]p_{x}-p_{y}[/tex] is given by [tex](\hat{p}_{x}-\hat{p}_{y})\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_{x}(1-\hat{p}_{x})}{n_{x}}+\frac{\hat{p}_{y}(1-\hat{p}_{y})}{n_{y}}}[/tex]. [tex]\hat{p}_{x}=43/56 = 0.7679[/tex] and [tex]\hat{p}_{y}=25/41=0.6098[/tex]. Because we want a 99% confidence interval for the difference [tex]p_{x}-p_{y}[/tex], we have that [tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5758[/tex] (The area above 2.5758 and below the curve of the stardard normal density is 0.005) and the confidence interval is [tex](0.7679-0.6098)\pm (2.5758)\sqrt{\frac{0.7679(1-0.7679)}{56}+\frac{0.6098(1-0.6098)}{41}}[/tex] = (-0.0861, 0.4023).

This answer includes instructions for approximating a function using a third degree Taylor polynomial at a specific point, using Taylor's Inequality to estimate the accuracy of the approximation, and checking the results by graphing the function |Rn(x)|.

For part (a), to approximate the function f(x) = ln(1 + 2x) using a Taylor polynomial of degree 3 at a = 4, we first need to find the first 4 derivatives of the function evaluated at a=4. Substituting the values will give the Taylor polynomial - T3(x).

For part (b), Taylor's Inequality states that the remainder term Rn(x) is less than or equal to M|X-a|^(n+1) / (n+1)! where M is the max value of the |f^(n+1)(t)| where t lies in the interval between a and x. By substituting the values we can find |R3(x)|.

For the last part (c), by using graphing software, you can graph the function |Rn(x)| in the given interval to visually confirm your earlier result. These graphs will show the magnitude of the error.|

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Answer:

x=-3+4t\\

y =5-7t

Step-by-step explanation:

we are given that when two point (x1,y1) and (x2,y2) are joined the parametric equation representing the line segment would be

[tex]x = x_1 + (x_2 - x_1)t y = y_1 + (y_2 - y_1)t, 0 \leq ≤t \leq ≤ 1[/tex]

Our points given are

[tex](-3, 5) to (1, -2)[/tex]

So substitute the values to get

[tex]x=-3+(1+3)t =-3+4t\\y = 5+(-2-3)t =5-7t[/tex]

Hence parametric equations for the line segment joining the points

(-3,5) and (1,-2) is

[tex]x=-3+4t\\y =5-7t[/tex]

Final answer:

The parametric equations for the line segment from (-3, 5) to (1, -2) are x = -3 + 4t and y = 5 - 7t for 0 ≤ t ≤ 1.

Explanation:

The question asks for parametric equations to describe the line segment joining the points P1(-3, 5) and P2(1, -2). According to the formula provided, we have:

x = x1 + (x2 – x1)t

y = y1 + (y2 – y1)t

Substituting the values of P1 and P2 into the equations:

x = -3 + (1 – (-3))t = -3 + 4t

y = 5 + (-2 – 5)t = 5 - 7t

This means for 0 ≤ t ≤ 1, the parametric equations representing the line segment from (-3, 5) to (1, -2) are:

x = -3 + 4t

y = 5 - 7t

Answer:

To conduct the hypothesis test I would use the standard normal distribution to calculate the critical value and the p-value.

Step-by-step explanation:

To conduct the hypothesis test I would use the standard normal distribution because there is a large sample size of n = 125 households. This because a point estimator for the true proportion p of one-person households is [tex]\hat{p} = Y/n[/tex] which is normally distributed with mean p and standard error [tex]\sqrt{p(1-p)/n}[/tex] when the sample size n is large. Here Y is the random variable that represents the number of one-person households observed. Then the test statistic is [tex]Z = \frac{\hat{p}-0.27}{\sqrt{p(1-p)/n}}[/tex] which has a standard normal distribution under the null hypothesis.

Answer: Slope 1/2

y-int -5

graph (0,-5),(1,-9/2)

Step-by-step explanation:

Answer:

The length of the person’s shadow is 5.7ft

Explanation:

Length of the flagpole =a= 35ft

Length of the shadow of the flagpole= b=50ft

Length of the person=c= 4ft

Suppose the length of the person’s shadow is=d

According to the  rules of trigonometry

[tex]\frac{\text { Length of the flagpole }}{\text { Length of the shadow of the flagpole }}=\frac{\text { Length of the person }}{\text { Length of the person's shadow }}[/tex]

[tex]\frac{a}{b}=\frac{c}{d}[/tex]

[tex]\frac{35}{50}=\frac{4}{d}[/tex]

35d=200

d=[tex]\frac{200}{35}[/tex]

d=5.7ft  

Hence, The length of the person’s shadow is 5.7ft.

Answer:

;-; I belive it's the 1st or 2nd one

Step-by-step explanation:

hope this helps you.

ask me for explanation if you want and I'll try to explane it the best I Can.

(P.S;please mark me as brainlyest,ty in advaince)

Answer: t = 1.9287

Step-by-step explanation:

Let [tex]\mu[/tex] be the average number of pages should be sent by 2-day mail instead.

As per given we have,

[tex]H_0: \mu \leq7\\\\H_a:\mu>7[/tex]

Sample mean : [tex]\overline{x}=8.52[/tex]

Sample standard deviation : s=3.81

sample size : n= 24

Since , the sample size is less than 30 and populations standard deviation is unknown , so we use t-test.

The test statistic for population mean :-

[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

[tex]t=\dfrac{8.5-7}{\dfrac{3.81}{\sqrt{24}}}\\\\=\dfrac{1.5}{\dfrac{3.81}{4.8990}}\\\\=\dfrac{1.5}{0.777712993333}=1.92873208093\approx1.9287[/tex]

Hence, the test statistics : t = 1.9287

Answer:

B and D

Step-by-step explanation:

sin x cos x = 1/4

Multiply both sides by 2:

2 sin x cos x = 1/2

Use double angle formula:

sin(2x) = 1/2

Solve:

2x = π/6 + 2nπ or 2x = 5π/6 + 2nπ

x = π/12 + nπ or x = 5π/12 + nπ

Answer:

AAS method can be used to prove that the two triangles are congruent.

Step-by-step explanation:

According to the question for the two triangles one pair of opposite angles are equal. One another pair of angles are equal for the two and one pair of sides are also equal of the two.

Hence, the two given triangles are congruent by AAS rule.

Hence, AAS method can be used to prove that the two triangles are congruent.

Answer: 95% confidence interval would be (175.04,194.96).

Step-by-step explanation:

Since we have given that

Mean = 185 mg

Standard deviation = 17.6 mg

At 95% confidence level, z = 1.96

So, Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=185\pm 1.96\times \dfrac{17.6}{\sqrt{12}}\\\\=185\pm 9.958\\\\=(185-9.958,185+9.958)\\\\=(175.042,194.958)\\\\=(175.04,194.96)[/tex]

Hence, 95% confidence interval would be (175.04,194.96).

Using the provided values and steps for constructing a 95% confidence interval, we estimate, with 95% confidence, that the true mean cholesterol content of all such eggs lies between 146.5 and 223.5 milligrams.

To construct a 95% confidence interval for the mean cholesterol content of all such eggs, we will use the provided sample mean (185 milligrams), standard error (17.6 milligrams), and the fact that the sample size (12 eggs) is relatively small, so we use a t-distribution.

Firstly, we need to find the t-score for a 95% confidence level. Since degrees of freedom (df) is n-1 -> 12-1=11, and at a 95% confidence level, the t-score (from t-distribution table) is approximately 2.201 for a two-tailed test.

Next, we use the formula for confidence interval:

Calculating these gives:

So, we estimate with 95 percent confidence that the true mean cholesterol content of all such eggs is between 146.5 and 223.5 milligrams.

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Answer:

Part A)  Not collide

Part B)  k = 4

Part C)  Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

Position of particle A:-

[tex]x(t)=5t-5[/tex]

[tex]y(t)=2t-k[/tex]

Position of particles B:-

[tex]x(t)=4t[/tex]

[tex]y(t)=t^2-2t+1[/tex]

Part A)  For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, [tex](4t,t^2-2t-1)[/tex]

If both collides then x and y coordinate must be same

Therefore,

5t - 5 = 4t    

       t = 5

[tex]2t+6=t^2-2t-1[/tex]

[tex]t^2-4t-7=0[/tex]

[tex]t=-1.3,5.3[/tex]

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

5t - 5 = 4t    

       t = 5

[tex]2t-k=t^2-2t-1[/tex]

Put t = 5

[tex]10-k=25-10-1[/tex]

[tex]k=4[/tex]

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, [tex]\dfrac{dA}{dt}[/tex]

[tex]\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4[/tex]

Speed of particle B, [tex]\dfrac{dB}{dt}[/tex]

[tex]\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}[/tex]

At t = 5

[tex]\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2[/tex]

Hence, Particle B moves faster than particle A

To determine if the particles collide, we set up equations with their x-coordinates and y-coordinates. Part (a) asks if they collide when k = -6. Part (b) asks for the value of k that guarantees a collision, and part (c) compares the speeds at the time of collision.

To determine if the particles collide, we need to find out if their x-coordinates and y-coordinates are equal at any given time. Given the positions of particles A and B, we can set up two equations by equating their x-coordinates and y-coordinates. For part (a), when k = -6 we can solve the equations to find if they intersect. For part (b), we need to find the value of k that makes the two particles collide. Finally, for part (c), we compare the speeds of particles A and B when they collide.

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Answer:

Lowest acceptable score = 121.3

Step-by-step explanation:

Mean test score (μ) = 115

Standard deviation (σ) = 12

The z-score for any given test score 'X' is defined as:  

[tex]z=\frac{X-\mu}{\sigma}[/tex]  

In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:

[tex]0.525=\frac{X-115}{12}\\X=121.3[/tex]

Answer:

6x^2 + 18x + 19.

Step-by-step explanation:

5x^2 + 6x - 17 + (x + 6)^2

= 5x^2 + 6x - 17 + x^2 + 12x + 36

=  6x^2 + 18x + 19.

Answer:

6x^2 + 18x + 19

Step-by-step explanation:

The standard form of a polynomial depends on the degree of the polynomial. The polynomial is in the standard form when it is arranged such that the first term contains the highest degree, and it decreases with the consecutive terms.

The square of the binomial is (x+6)^2

(x+6)^2 = (x+6)(x+6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36

The sum of the two polynomials will be 5x^2+6x−17 + x^2 + 12x + 36

Collecting like terms,

5x^2 + x^2 + 6x + 12x + 36 -17

= 6x^2 + 18x + 19

One sample t-test

Step-by-step explanation:

In this statistical test, you will be able to test if a sample mean, significantly differs from a hypothesized value.Here you can test if the average swimming velocity differs significantly from an identified value in the hypothesis.Then you can conclude whether the group of 18 fish or that of 21 fish has a significantly higher or lower  mean velocity than the one in the hypothesis.

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Answer:

z=0.77

Step-by-step explanation:

If 22% of the area under the standard normal curve lies to the right of z, 78% lies to the left.

According to a one-sided z-score table:

If z-score = 0.77 the area under the curve is 77.94%

If z-score = 0.78 the area under the curve is 78.23%

Since only two decimal plates are required, there is no need to actually interpolate the values, just verify which one is closer to 78%

Since 77.94% is closer, a z-score of 0.77 represents a 22% area under the standard normal curve to the right of z.

To find z such that 22% of the area under the standard normal curve lies to the right of z, the value of z is approximately 0.81.

To find the value of z such that 22% of the area under the standard normal curve lies to the right of z, we can use a z-table or a calculator. First, we need to find the area under the curve to the left of z, which is 1 - 0.22 = 0.78. We can then look up the z-score that corresponds to this area in the table. The z-score is approximately 0.81. Therefore, the value of z is approximately 0.81.

Answer:

5

Step-by-step explanation:

27-22= 22-17 = 17-12 =12-7=d

d= 5

The common difference of the arithmetic sequence 7, 12, 17, 22, 27,... is 5, which is the constant difference between consecutive terms of the sequence.

The sequence given is an arithmetic sequence, which is a sequence of numbers where the difference between consecutive terms is constant. To find the common difference in the arithmetic sequence 7, 12, 17, 22, 27,..., we subtract any term from the preceding term in the sequence. Let's perform this calculation using the first two terms:

Thus, the common difference is 5. We can confirm this by checking the difference between other consecutive terms:

All differences are the same, confirming that the common difference of the sequence is indeed 5.

Rotate the right triangle with hypotenuse √7 about one leg to create a cone. Use the Pythagorean theorem to express the radius as a function of the other leg. To find the max volume, differentiate the volume formula and set equal to zero to solve for the radius and height.

The problem presents a right triangle with hypotenuse of √7 meters. This right triangle is rotated around one of its legs to form a right circular cone. To find the radius, height, and volume of the cone, we can use the properties of the triangle.  

Imagine that the triangle is rotated about the shorter leg. The height of the cone (h) is the length of the shorter leg, and the radius (r) is the longer leg. By the Pythagorean theorem, a² + b² = c², where c is the hypotenuse, a and b represent the legs of the triangle. Assume that a is the longer leg and b is the shorter leg.

From the Pythagorean theorem we get a² = (√7)² - b² = 7 - b² and so a = √(7 - b²).

The volume of the cone V = (1/3)πr²h = (1/3)π(√(7 - b²))²b = (1/3)π(7 - b²)b. Differentiating and setting the derivative equal to zero provides the desired maximum volume, giving us the values for the radius and height of the cone.

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Answer:

Step-by-step explanation:

She uses 1/4 ounces of green paint to draw to draw a green line on her painting. This means if she draws 2 green lines on her painting, she will require ( 1/4 + 1/4 ) ounces of green paint. This means that if she draws a total of 7 green lines, the correct options would be the options that apply and they are just two viz:

1/4+1/4+1/4+1/4+1/4+1/4+1/4 or

7/4 because they are both the same. Adding each term in the longer oftion gives 7/4

Answer:

c and d

Step-by-step explanation:

The setup for the surface integral is achieved by using the parameterized surface and the given function along with the norm of the cross product of partial derivatives of the parameterized surface. The integral is evaluated to represent the net flux through the surface that denotes the volume enclosed by the parameterized surface.

The surface integral of a given function can be calculated using the parameterization of the surface and the given function. In this case, our surface is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩ and our function is f(x,y,z)=x²+y²+z². We first need to calculate the partial derivatives ∂r/∂u and ∂r/∂v, and then cross them to find the norm of the cross product, which gives us the differential area element for the parameterized surface. With these, we can set up the surface integral as follows: ∫σ∫f(x(u,v),y(u,v),z(u,v))∥∥∥∂r∂u×∂r∂v∥∥∥dA = ∫0 to 2π/3 ∫0 to 7 [u²cos²(3v)+u²sin²(3v)+v²]∥∂r/∂u×∂r/∂v∥ du dv.

Please note that the evaluated result of the integral provides the net flux through the rectangular surface which represents the volume enclosed by the parameterized surface.

The above set up integral needs to be solved to find the exact value which is not required in this scenario. However, the integral can be solved using suitable integral calculus methods if needed.

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A function is a relation that has one output for a given input.

For the first one, there is no one x value with two or more y values so it is a function.

The second example is also a function because a certain value can only have one cube root.

For problem number 3 input "-3" for every instance of x in h(x).

So, h(-3)=2(3^2)-1= 17

Answer:

0.212 is the probability that a randomly selected tire will have a depth less than 0.70 mm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  0.98 mm

Standard Deviation, σ =  0.35 mm

We are given that the distribution of tire tread is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(depth less than 0.70 mm)

P(x < 0.70)

[tex]P( x < 0.70) = P( z < \displaystyle\frac{0.70 - 0.98}{0.35}) = P(z < -0.8)[/tex]

Calculating from normal z table, we have:

[tex]P(z<-0.8) = 0.212[/tex]

[tex]P(x < 0.70) = 0.212 = 21.2\%[/tex]

Thus, this event is not unusual and will  not warrant a refund.

Answer:

82 apartments should be rented.

Maximum profit realized will be $30964.

Step-by-step explanation:

Monthly profit realized from renting out x apartments is modeled by

P(x) = -11x² + 1804x - 43000

To maximize the profit we will take the derivative of the function P(x) with respect to x and equate it to zero.

P'(x) = [tex]\frac{d}{dx}(-11x^{2}+1804x-43000)[/tex]

       = -22x + 1804

For P'(x) = 0,

-22x + 1804 = 0

22x = 1804

x = 82

Now we will take second derivative,

P"(x) = -22

(-) negative value of second derivative confirms that profit will be maximum if 82 apartments are rented.

For maximum profit,

P(82) = -11(82)² + 1804(82) - 43000

        = -73964 + 147928 - 43000

        = $30964

Therefore, maximum monthly profit will be $30964.

The number of units that should be rented out to maximize the monthly rental profit is 82 units.

The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

To find the number of units that should be rented out to maximize the monthly rental profit and the maximum monthly profit, we need to analyze the given profit function:

[tex]\[ P(x) = -11x^2 + 1804x - 43,000 \][/tex]

This is a quadratic function of the form [tex]\( P(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -11 \), \( b = 1804 \)[/tex], and c = -43,000 .

For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] with a < 0  (which opens downwards), the maximum value occurs at the vertex. The x-coordinate of the vertex for the function P(x)  is given by:

[tex]\[ x = -\frac{b}{2a} \][/tex]

Substituting a = -11  and b = 1804 :

[tex]\[ x = -\frac{1804}{2(-11)} = \frac{1804}{22} = 82 \][/tex]

Therefore, the number of units that should be rented out to maximize the profit is x = 82 .

Next, we calculate the maximum monthly profit by substituting x = 82  back into the profit function:

[tex]\[ P(82) = -11(82)^2 + 1804(82) - 43,000 \][/tex]

Calculating step by step:

[tex]\[ 82^2 = 6724 \]\[ -11(6724) = -73,964 \]\[ 1804(82) = 147,928 \]\[ P(82) = -73,964 + 147,928 - 43,000 \]\[ P(82) = 30,964 \][/tex]

So, the maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

In summary:

- The number of units that should be rented out to maximize the monthly rental profit is 82 units.

- The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

Answer:

$302,500

Step-by-step explanation:

If cost (C) = $7, then Sales (S) = 37,500 units

If cost (C) = $15, then Sales (S) = 17,500 units

The slope of the linear relationship between units sold and cost is:

[tex]m=\frac{37,500-17,500}{7-15}\\m= -2,500[/tex]

The linear equation that describes this relationship is:

[tex]s-s_0 =m(c-c_0)\\s-17500 =-2500(c-15)\\s(c)=-2500c + 55,000[/tex]

The revenue function is given by:

[tex]R(c) = c*s(c)\\R(c)=-2500c^2 + 55,000c[/tex]

The cost at which the derivative of the revenue equals zero is the cost that yields the maximum revenue.

[tex]\frac{d(R(c))}{dc}=0 =-5000c + 55,000\\c=\$11[/tex]

The optimal cost is $11, therefore, the maximum revenue is:

[tex]R(11)=-2500*11^2 + 55,000*11\\R(11)=\$ 302,500[/tex]

On average, when placing a $1 bet on five numbers in roulette, you would expect to lose about 53 cents due to the game's probability structure.

In order to calculate the expected winnings in a roulette game, consider a $1 bet on five numbers. The roulette wheel has 38 possibilities (1 to 36, 0, and 00). So, the chance that your bet will win is 5 out of 38. If it wins, the game pays off 6 to 1, which means you get your $1 bet back and win $6 additional, for a total of $7. The expected value of this bet can be calculated as follows: (probability of winning * amount won if bet is successful) - (probability of losing * amount lost if bet is not successful). In other words, E(X) = [5/38 * $7] - [33/38 * $1]. This equates to an expected value of -$0.53 (rounded to the nearest cent). So, on average, with each $1 bet on five numbers, you would expect to lose about 53 cents.

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In the game of roulette with 38 outcomes, if you place a $1 bet on five numbers, your expected winnings are -$0.08, which means you lose 8 cents per game on average.

To calculate the expected winnings of a bet in this roulette scenario, you need to know the probability of winning and the associated payout, and also the probability of losing. In a roulette wheel with numbers 1-36 plus 0 and 00, there are a total of 38 possible outcomes. When you place a 5-number bet, the probability of winning is 5 out of 38 or approximately 0.1316, and the probability of losing is 33 out of 38 or approximately 0.8684.

Now we add the expected winnings and losses. The expected gain from a win is the probability of winning times the payout, which is $6 in this case. So, 0.1316 * $6 = $0.79, rounded to the nearest cent. The expected loss from a bet is the probability of loss times the amount of the bet which is $1. So, 0.8684 * -$1 = -$0.87, rounded to the nearest cent. Total expected value or earnings from a single $1 bet on five numbers would be the sum of expected gains and losses, or $0.79 - $0.87 = -$0.08.

So, with a $1 bet on five numbers on this roulette wheel, you can expect, on average, to lose 8 cents per game. This negative figure indicates that this is not a good bet if you're expecting to make money on average.

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Answer: The lower bound of confidence interval would be 0.116.

Step-by-step explanation:

Since we have given that

p = 13.2%= 0.132

n = 1105

At 90%  confidence,

z = 1.645

So, Margin of error would be

[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.132\times 0.868}{1152}}}\\\\=0.0164[/tex]

So, the lower bound of the confidence interval would be

[tex]p-\text{margin of error}\\\\=0.132-0.0164\\\\=0.116[/tex]

Hence, the lower bound of confidence interval would be 0.116.

The confidence interval for percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.

To estimate the percentage of U.S. adults who were victims of crime at the 90% confidence level, we can use the formula for a confidence interval:

[tex]CI = p +/- Z * \sqrt{((p * (1-p)) / n)[/tex]

where CI is the confidence interval, p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size. In this case, p = 0.132, Z = 1.645 (corresponding to a 90% confidence level), and n = 1105. Plugging in these values, we can calculate the confidence interval as:

[tex]CI = 0.132 +/- 1.645 * \sqrt((0.132 * (1-0.132)) / 1105)[/tex]

Simplifying the expression gives us:

CI = 0.132 ± 0.018

Therefore, the confidence interval for the percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.

Answer:

[tex]P(\bar X>308)=1-0.0721=0.928[/tex]

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the GRE score at the University of Pennsylvania for the incoming class of 2016-2017, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=311,\sigma=13)[/tex]  

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case  [tex]\bar X \sim N(311,\frac{13}{\sqrt{40}})[/tex]

2) Calculate the probability

We want this probability:

[tex]P(\bar X>308)=1-P(\bar X<308)[/tex]

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(\bar X >308)=1-P(Z<\frac{308-311}{\frac{13}{\sqrt{40}}})=1-P(Z<-1.46)[/tex]

[tex]P(\bar X>308)=1-0.0721=0.9279[/tex]  and rounded would be 0.928

Answer:

Is plausible that the successive throws are independent

Step-by-step explanation:

1) Table with info given

The observed values are given by the following table

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First shot          Made          Second shot missed           Total

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Made                  152                       33                                185

Missed                37                         8                                  45

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Total                    189                       41                                230

2) Calculations and test

We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:

Null hypothesis: Independence between two successive free throws

Alternative hypothesis: No Independence between two successive free throws

_____________________________________________________

First shot                    Made                                    Second shot missed

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Made                  189(185)/230=152.0217                41(185)/230=32.9783

Missed                189(45)/230=36.9783                  41(45)/230=8.0217

_____________________________________________________

On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.

3) Calculate the chi square statistic

The statistic for this case is given by:

[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]

Where O represent the observed values and E the expected values. Replacing the values that we got we have this

[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]

Now with the calculated value we can find the degrees of freedom

[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.

Now we can calculate the p value

[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]

On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.

Answer:

Step-by-step explanation:

To find median and mode for

a) In a uniform distribution median would be

(a+b)/2 and mode = any value

b) X is N

we know that in a normal bell shaped curve, mean = median = mode

Hence mode = median = [tex]\mu[/tex]

c) Exponential with parameter lambda

Median = [tex]\frac{ln2}{\lambda }[/tex]

Mode =0

The median of a distribution is the middle value while the mode is the highest occuring value

The median (M) of a uniform distribution is:

[tex]M = \frac{a +b}2[/tex]

A uniform distribution has no mode

For a normal distribution with the given parameters, we have:

Median = Mean = Mode = μ

Hence, the median and the mode are μ

For an exponential distribution with the given parameter, we have:

[tex]Median = \frac{\ln 2}{\lambda}[/tex]

The mode of an exponential distribution is 0

Read more about median and mode at:

https://brainly.com/question/14532771