A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of 215 cars owned by students had an average age of 7.41 years. A sample of 252 cars owned by faculty had an average age of 6.9 years. Assume that the population standard deviation for cars owned by students is 3.72 years, while the population standard deviation for cars owned by faculty is 2.26 years. Determine the 98%98% confidence interval for the difference between the true mean ages for cars owned by students and faculty.
Step 1 of 3: Find the point estimate for the true difference between the population means.
Step 2 of 3: Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places
. Step 3 of 3: Construct the 98% confidence interval. Round your answers to two decimal places.

Given:

[tex]\text{AOC}[/tex] and [tex]\text{BOD}[/tex] are diameters of a circle and has center [tex]\text{O}[/tex].

To Find:

[tex]\Delta\text{ABD}[/tex] and [tex]\Delta\text{DCA}[/tex] are congruent by [tex]\text{RHS}[/tex].

Solution:

It is given that [tex]\text{AOC}[/tex] and [tex]\text{BOD}[/tex] are diameters of a circle.

[tex]\rightarrow \text{BD} = \text{CA}[/tex] [diameters of the circle]

[tex]\rightarrow \angle\text{BAD} = \angle\text{CDA}[/tex] [angles in semicircle is 90°]

[tex]\rightarrow \text{AD} = \text{AD}[/tex] [common in both the triangles]

[tex]\rightarrow \Delta\text{ABD} \cong \Delta\text{DCA}[/tex] [using RHS congruence criteria]

Hence, proved [tex]\Delta\bold{ABD} \cong \Delta\bold{DCA}[/tex] by [tex]\bold{RHS}[/tex] congruency criteria.

When conducting a hypothesis test, a(an) **b. t-statistic** is more appropriate than a z-score when you don't know the population variance or the population standard deviation. The t-statistic takes into account the sample size and is better suited for situations where population parameters are unknown.

When conducting a hypothesis test, a t-statistic is more appropriate than a 2-score when you don't know the population variance or the population standard deviation. The t-statistic is used to test hypotheses about population means when the sample size is small or when the population standard deviation is unknown.

The t-statistic is calculated by dividing the difference between the sample mean and the hypothesized population mean by the standard error of the mean, which takes into account the variability of the sample.

The t-statistic is compared to a critical value from a t-distribution with n-1 degrees of freedom, where n is the sample size. The level of significance, or alpha value, is also used to determine the critical value. Sample variance and Cohen's d are other statistical measures used in hypothesis testing but are not specifically related to the use of t-statistics.

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Option C, "A study shows that the average student leaves a four-year college with a student loan debt of more than $30,000", involves descriptive statistics.

The area of statistics known as descriptive statistics deals with the gathering, organizing, organizing, analyzing, interpreting, and presenting of data. It summarizes and describes the main features of a dataset, including measures of central tendency (such as mean, median, and mode) and measures of variability (such as range, standard deviation, and variance). Option C presents a descriptive statistic (the average student loan debt) that summarizes a larger dataset, making it the correct answer.

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The table which contains only values that satisfy the equation of line defined as y = 0. 5x + 14, ( linear equation) is present in option(c). So, option(c) is right one.

We have a equation of line, y = 0.5x + 14, --(1) which is a equation of line . We have to recognise the table which satisfy the above line equation. The values are called roots of the equation. A value that is a solution of an equation is said to satisfy the equation, and the solutions of an equation create its solution set. The above table consists values of x and y, so we check which set of values form solution set of equation (1). Let x = 0 => y = 14 so, ( 0, 14) is solution of equation (1). Similarly, when x = 5

=> y = 5× 0.5 + 14 = 16.5

Similarly, we can check other point values. The table present in option (c) is correct answer.

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Complete question:

The above figure complete the question.

the dimensions of the rectangle are L = 15 inches and W = 15 inches, and the minimum perimeter is:    P = 2L + 2W = 60 inches.

Let the length and width of the rectangle be L and W, respectively, so that the area of the rectangle is A = LW = 225. We want to find the dimensions of the rectangle with minimum perimeter P = 2L + 2W, and then find the minimum perimeter.

Using the given area, we can solve for one of the variables in terms of the other:

L = 225/W

Substituting this expression for L into the expression for the perimeter, we get:

P = 2(225/W) + 2W

Taking the derivative of P with respect to W and setting it equal to zero to find the minimum, we get:

[tex]dP/dW = -450/W^2 + 2 = 0[/tex]

Solving for W, we get:

W^2 = 225

Since W must be positive (it is a length), we take the positive square root:

W = 15

Substituting this value of W back into the expression for L, we get:

L = 225/W = 15

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The population of Pennsylvania is: 1304503 people

Population density is calculated by taking the total area of a region in question and dividing it by the total number of people that live in that area. The result will give the average number of inhabitants per square kilometre, mile, acre, meter, etc.

The parameters given are:

Population of colorado = 5,770,000 people

Area of colorado = 280 * 380

= 106,400 mi²

Population density here = 5,770,000/106,400

54.23 people per mi²

Area of Pennsylvania = 283 * 170

= 48110 mi²

Thus:

Population of Pennsylvania/48110 mi² = 5 * 54.23 people per mi²

Population of Pennsylvania = 48110 * mi² * 5 * 54.23 people per mi²

= 1304503 people

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Answer:9.5

Step-by-step explanation:

Answer: 8

Step-by-step explanation:

The median of this data set is 8. If you cross one number from both sides at the same time, you will eventually come to the middle of the data set, which is 8.

How much does it take to cover the cylinder? is different from the rest, as it refers to the surface area of the cylinder, not its volume or capacity.

The term "cover" usually means to place something over the top or on the surface of something else, such as a lid covering a container.

In the context of a cylinder, "covering" would typically refer to finding the surface area of the cylinder, which includes both the top and bottom circles as well as the curved lateral surface.

In contrast, the other statements are related to the volume or capacity of the cylinder, which refers to how much space is contained inside the cylinder.

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Therefore, if Isabel plans to drive inequality more than 25 miles, Company B will be more expensive than Company A. If she plans to drive 25 miles or less, Company A will be more expensive.

Let's start by setting up an inequality to represent the mileages for which Company A charges less than Company B.

For Company A, the cost is a flat fee of $90, regardless of the number of miles driven.

For Company B, the cost depends on the number of miles driven. The initial fee is $75, and then there is an additional charge of $0.60 for every mile driven. So, the total cost for Company B can be represented by the equation:

Cost(B) = 0.60m + 75

here m is the number of miles driven.

We want to find the mileages for which Company A charges less than Company B. In other words, we want to find the values of m for which:

Cost(A) < Cost(B)

Substituting in the expressions for the costs, we get:

90 < 0.60m + 75

Simplifying and solving for m, we get:

15 < 0.60m

25 < m

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The three options with dimensions resulting in a surface area greater than 140 square feet are:

To determine whether the dimensions of a rectangular prism result in a surface area greater than 140 square feet, we can use the formula for the surface area of a rectangular prism:

Surface Area = 2lw + 2lh + 2wh

    where l, w, and h are the length, width, and height of the rectangular prism, respectively.

Option 1: 6 feet by 2 feet by 3 feet

Surface Area = 2(6)(2) + 2(6)(3) + 2(2)(3) = 24 + 36 + 12 = 72 square feet

This option does not have a surface area greater than 140 square feet.

Option 2: 6 feet by 5 feet by 4 feet

Surface Area = 2(6)(5) + 2(6)(4) + 2(5)(4) = 60 + 48 + 40 = 148 square feet

This option has a surface area greater than 140 square feet.

Option 3: 7 feet by 6 feet by 4 feet

Surface Area = 2(7)(6) + 2(7)(4) + 2(6)(4) = 84 + 56 + 48 = 188 square feet

This option has a surface area greater than 140 square feet.

Option 4: 8 feet by 3 feet by 7 feet

Surface Area = 2(8)(3) + 2(8)(7) + 2(3)(7) = 48 + 112 + 42 = 202 square feet

This option has a surface area greater than 140 square feet.

Option 5: 8 feet by 4 feet by 3 feet

Surface Area = 2(8)(4) + 2(8)(3) + 2(4)(3) = 64 + 48 + 24 = 136 square feet

This option does not have a surface area greater than 140 square feet.

Therefore, the three options with dimensions resulting in a surface area greater than 140 square feet are:

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Answer:

Part A:

Sunday newspaper: Start at $36 and add $4 for each paper delivered.

Saturday newspaper: Start at $12 and add $2.50 for each paper delivered.

Part B:

To calculate Stevic's total earnings after delivering 15 papers on each day:

Earnings from Sunday papers = $36 + ($4 x 15) = $96

Earnings from Saturday papers = $12 + ($2.50 x 15) = $49.50

Total earnings = Earnings from Sunday papers + Earnings from Saturday papers

Total earnings = $96 + $49.50

Total earnings = $145.50

Therefore, Stevic's total earnings after delivering 15 papers on each day is $145.50.

Step-by-step explanation:

The risk due to the hazard of exposure to radioactive chemicals is 6 cancer-deaths per year.  The probability of exposure is approximately 0.002. The event tree exposure with 1, 2, or 3 barriers failing, and no exposure if all 3 barriers work.

To calculate the probability of exposure and risk due to the hazard, we need to develop an event tree showing all the branches and outcomes.

The event tree for this scenario would look like this:

Barrier 1 fails (0.001 probability) -> Exposure -> 3000 cancer-deaths per year
Barrier 2 fails (0.001 probability) -> Barrier 1 works -> Exposure -> 3000 cancer-deaths per year
Barrier 3 fails (0.001 probability) -> Barrier 2 works -> Barrier 1 works -> Exposure -> 3000 cancer-deaths per year
All 3 barriers work -> No exposure -> No consequence

                               Start

                                |

                            Barrier 1

                           /     |     \

                    Fail (0.001)  |   Pass (0.999)

                      |            |

           Exposure    Barrier 2

(3000 cancer-deaths) /    |     \

                                    /     |     \

                      Barrier 2  |   Barrier 3

                     Fail (0.001)|   Pass (0.999)

                       |         |

                   Exposure  No exposure

            (3000 cancer-deaths)     |

                                   |

                              Barrier 3

                             Fail (0.001)

                               |

                            Exposure

                     (3000 cancer-deaths)

From this event tree, we can see that there are 4 possible outcomes: exposure with 1, 2, or 3 barriers failing, and no exposure if all 3 barriers work.

The probability of exposure can be calculated by adding up the probabilities of each branch that leads to exposure:
0.001 + (0.001 x 0.999) + (0.001 x 0.999 x 0.999) = 0.001997

Therefore, the probability of exposure is approximately 0.002 (or 0.2%).

To calculate the risk, we need to multiply the probability of exposure by the consequence:
0.002 x 3000 = 6

Therefore, the risk due to the hazard of exposure to radioactive chemicals is 6 cancer-deaths per year. However, it is important to continue to monitor and maintain these barriers to ensure their effectiveness and minimize the risk of exposure.

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Part A - The pH of the buffer solution is 6.37.

Net ionic equation is [tex]H_2CO_3[/tex] + [tex]OH^-[/tex] → [tex]HCO^{3-}[/tex] + [tex]H_2O[/tex]

Part B - The pH of the buffer solution is 6.38.

Net ionic equation is [tex]H_2CO_3[/tex] + [tex]I^-[/tex] → [tex]HCO^{3-}[/tex] + [tex]H_3O^+[/tex]

Part A: To find the pH of the buffer solution, we first need to calculate the pKa of the weak acid. The pKa is -log(Ka1), so pKa1 = -log(4.20 x [tex]10^{-7}[/tex]) = 6.38.

Next, we can use the Henderson-Hasselbalch equation to find the pH: pH = pKa1 + log([[tex]A^-[/tex]]/[HA]).

Plugging in the values for the buffer solution, we get pH = 6.38 + log(0.304/0.304) = 6.38. Therefore, the pH of the buffer solution is 6.38.

The net ionic equation for the reaction when 0.088 mol KOH is added to 1.00 L of the buffer solution is:

[tex]H^+[/tex] + [tex]OH^-[/tex] → [tex]H_2O[/tex]

Part B: Similar to Part A, we first need to calculate the pKa of the weak acid. pKa1 = -log(4.20 x [tex]10^{-7}[/tex]) = 6.38.

Then, we can use the Henderson-Hasselbalch equation to find the pH: pH = pKa1 + log([A-]/[HA]).

Plugging in the values for the buffer solution, we get pH = 6.38 + log(0.311/0.311) = 6.38. Therefore, the pH of the buffer solution is 6.38.

The net ionic equation for the reaction when 0.089 mol HI is added to 1.00 L of the buffer solution is:

[tex]H_3O^+[/tex] + [tex]I^-[/tex] → HI + H2O

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The quadratic equation with zeros at -6 and 2 is y² + 4y - 12 = 0. This is in standard form, which is ax² + bx + c = 0, with a = 1, b = 4, and c = -12.

To find the quadratic equation with zeros at -6 and 2, we can start by using the fact that if a quadratic equation has roots x₁ and x₂, then it can be written in the form

(y - x₁)(y - x₂) = 0

where y is the variable in the quadratic equation.

Substituting the given values of the zeros, we get

(y - (-6))(y - 2) = 0

Simplifying this expression, we get

(y + 6)(y - 2) = 0

Expanding this expression, we get

y² - 2y + 6y - 12 = 0

Simplifying this expression further, we get

y² + 4y - 12 = 0

So the quadratic equation with zeros at -6 and 2 is

y² + 4y - 12 = 0

This is the standard form of a quadratic equation, which is

ax² + bx + c = 0

where a, b, and c are constants. In this case, a = 1, b = 4, and c = -12.

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I)

[tex]~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{9}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 +1}{2}~~~ ,~~~ \cfrac{ 9 +5}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ 14 }{2} \right)\implies (2~~,~~7)[/tex]

II)

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the line AB

[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{9}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{9}-\stackrel{y1}{5}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{1}}} \implies \cfrac{ 4 }{ 2 } \implies 2 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ 2 \implies \cfrac{2}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{2} }}[/tex]

so we're really looking for the equation of a line whose slope is -1/2 and it passes through (2 , 7)

[tex](\stackrel{x_1}{2}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7}=\stackrel{m}{- \cfrac{1}{2}}(x-\stackrel{x_1}{2}) \\\\\\ y-7=- \cfrac{1}{2}x+1\implies {\Large \begin{array}{llll} y=- \cfrac{1}{2}x+8 \end{array}}[/tex]

The solution to the inequality is:

x ∈ (26/33, ∞)

i) We can simplify the left-hand side of the inequality as follows:

2|2x + 71| + 2 ≤ 24

2|2x + 71| ≤ 22

|2x + 71| ≤ 11

Next, we can split this into two separate inequalities, depending on the sign of (2x + 71):

2x + 71 ≤ 11

2x ≤ -60

x ≤ -30

or

2x + 71 ≥ -11

2x ≥ -82

x ≥ -41

Therefore, the solution to the inequality is:

x ∈ (-∞, -30] ∪ [-41, ∞)

ii) We can solve for x as follows:

33x - 2 > 24

33x > 26

x > 26/33

Therefore, the solution to the inequality is:

x ∈ (26/33, ∞)

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The probability of the offspring being brown with long fur is 25%.

To determine the probability of offspring being brown with long fur from a cross between a brown, long-furred rabbit (Bbhh) and a white, short-furred rabbit (bbHh), we will use the terms dominant, recessive, and independent assortment.

Step 1: Set up the Punnett squares for each trait separately.
For fur color (B and b alleles):
Bb (brown, long-furred rabbit)×bb (white, short-furred rabbit)
Resulting in offspring genotypes:
Bb (brown fur)
Bb (brown fur)
bb (white fur)
bb (white fur)

For fur length (H and h alleles):
hh (brown, long-furred rabbit)×Hh (white, short-furred rabbit)
Resulting in offspring genotypes:
Hh (short fur)
Hh (short fur)
hh (long fur)
hh (long fur)

Step 2: Calculate the probabilities for each trait.
For brown fur: 2 out of 4 (50%)
For long fur: 2 out of 4 (50%)

Step 3: Calculate the combined probability.
Since both the B and H traits assort independently, we can multiply the probabilities of each trait occurring:
0.5 (brown fur) x 0.5 (long fur) = 0.25 (25%).

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We are 98% confident that the mean of the true population y lies between 10.8 and 13.4.

To construct the confidence interval, we first need to calculate the critical value of t using the given values of C and n. Since C = 0.98, we can find the level of significance as α = 1 - C = 0.02.

Using a t-table or calculator, the critical value of t for a two-tailed test with 14 degrees of freedom and

[tex]\frac{\alpha}{2} = 0.01[/tex] is approximately 2.977.

Next, we can calculate the sample standard deviation as s = σ/√n = [tex]\frac{0.95}{\sqrt{15}}= 0.245[/tex].

Then, we can use the formula for a confidence interval for the population mean using the t-distribution:
(y ± t)×[tex]\frac{s}{\sqrt{n}}[/tex]

Substituting the given values, we get:

(12.1 ± 2.977)×[tex]\frac{0.245}{\sqrt{15}}[/tex]

Simplifying and rounding to one decimal place, we get the confidence interval: (10.8, 13.4)

Therefore, we are 98% confident that the true population mean y lies between 10.8 and 13.4.

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The solution of the differential equation 4y'' − y = [tex] {e}^{x/2} [/tex] + 8 by variation of parameter method is y(x) = (15C - 16)[tex] {e}^{x/2} [/tex] + 15C' [tex] {ex}^{-x/2} [/tex]

To solve the differential equation by variation of parameters, we assume that the solution is of the form,

y(x) = u₁(x)y₁(x) + u₂(x)y₂(x), linearly independent solutions of the homogeneous equation are y₂(x) and y₂(x), and functions to be determined u₁(x) and u₂(x). The homogeneous equation associated with the given differential equation is,

4y'' - y = 0

The characteristic equation is,

4r² - 1 = 0 which has solutions r = ±1/2. Therefore, the general solution of the homogeneous equation is,

y(x) = C[tex] {e}^{x/2} [/tex] + C'[tex] {e}^{-x/2} [/tex]

C and C' are arbitrary constants.

Now, we need to find particular solutions of the non-homogeneous equation. We can guess that a particular solution has the form,

[tex] y_{p(x)} = A(x) {e}^{(x/2)} [/tex]

where A(x) is a function to be determined. We can find A(x) by substituting y_p(x) into the differential equation and solving for A(x). We have,

[tex] 4y_{p(x)} - y_{p(x)} = {e}^{(x/2)} +8 [/tex]

Differentiating twice and substituting these into the differential equation gives:

[tex]4( A"(x) + A'(x)) {e}^{2/y} 2 + \frac{A(x)}{4} - A(x) {e}^{(x/2)} = {e}^{(x/2)} + 8[/tex]

Simplifying and solving for A(x), we obtain,

A(x) = -16/15

Therefore, a particular solution of the differential equation is:

[tex]y_{p(x)} = \frac{ - 16}{15} {e}^{(x \div 2)} [/tex]

The general solution of the non-homogeneous equation is then,

y(x) = C[tex] {e}^{x/2} [/tex] + C'[tex] {e}^{-x/2} [/tex] [tex]\frac{ - 16}{15} {e}^{(x/2)} [/tex]

Simplifying and collecting terms, we get,

y(x) = (15C - 16)[tex] {e}^{x/2} [/tex] + 15C' [tex] {ex}^{-x/2} [/tex] ,where C and C' are arbitrary constants.

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Complete question - Solve the differential equation by variation of parameters. 4y'' − y = e^x/2 + 8.

The probability that a student studied for 5 hours is given as follows:

0.23.

A probability is calculated as the division of the desired number of outcomes by the total number of outcomes in the context of a problem/experiment.

The total number of students in this problem is given as follows:

1 + 3 + 2 + 5 + 9 + 7 + 3 = 30.

Out of those 30 students, 7 studied five hours, hence the probability is given as follows:

p = 7/30 = 0.23.

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Answer:

The correct answer is the option 3

To solve this problem, we need to find the z-score corresponding to the 14th percentile of the normal distribution. We can then use this z-score to find the corresponding value of K.

First, we find the z-score corresponding to the 14th percentile using a standard normal distribution table or calculator. The 14th percentile is equivalent to a cumulative probability of 0.14, which corresponds to a z-score of approximately -1.08.

Next, we use the formula z = (x - μ) / σ to find the corresponding value of K. Rearranging this formula, we get x = μ + z * σ. Plugging in the values we know, we get:

K = 11 + (-1.08) * 0.25
K = 10.73 cm

Therefore, there is a 14% chance that a randomly selected cylindrical metal bar has a length longer than 10.73 cm.

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a. The correlation matrix p =  [tex]\left[\begin{array}{ccc}1&1/3&2/5\\1/3&1&-3/5\\2/5&-3/5&1\end{array}\right][/tex]. b. The correlation between X1, and i/2X2 + 1/2X3 is 0.3.

a. The correlation matrix p can be calculated by dividing the covariance matrix by the product of the standard deviations of the variables:

p = [tex]\left[\begin{array}{ccc}1&1/3&2/5\\1/3&1&-3/5\\2/5&-3/5&1\end{array}\right][/tex]

b. To compute the correlation between X1 and i/2X2 + 1/2X3, we first need to calculate the standard deviations of the variables:

σ1 = sqrt(4) = 2

σ2 = sqrt(9) = 3

σ3 = sqrt(25) = 5

Then, we can calculate the covariance between X1 and i/2X2 + 1/2X3:

cov(X1, i/2X2 + 1/2X3) = cov(X1, i/2X2) + cov(X1, 1/2X3)

= i/2 * cov(X1, X2) + 1/2 * cov(X1, X3)

= i/2 * 1 + 1/2 * 2

= 1.5

Finally, we can compute the correlation using the formula:

corr(X1, i/2X2 + 1/2X3) = cov(X1, i/2X2 + 1/2X3) / (σ1 * σ2/2 + σ3/2)

= 1.5 / (2 * 3/2 + 5/2)

= 0.3

Therefore, the correlation between X1 and i/2X2 + 1/2X3 is 0.3.

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To determine the supremum and infimum of the given sets.

(a) The set {1/n: n ∈ N} consists of the reciprocals of positive integers. The smallest element in the set is 1, as it corresponds to n=1. The set has no largest element since it has an infinite number of elements getting smaller as n increases. Therefore, the infimum (greatest lower bound) of the set is 1, and there is no maximum. The supremum (least upper bound) of the set is not in the set itself, but it exists and equals 1.

(b) The set {z ∈ Q: 22 < 3} is an empty set since there is no rational number z that satisfies the condition 22 < 3. In this case, there is no supremum or infimum since the set has no elements. Consequently, there is no maximum or minimum value.

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Also, minimum observations for both data sets are same, however there is a difference in the maximum for the both data sets.

From the given boxplots, it is observed that the boxplot for the species BARN has more variation than the boxplot for the species OYST. The boxplot for the species OYST indicates that there is are some outliers present in the data, however the boxplot for the BARN species indicates that there are no any outliers present in the data. It is observed that the median for the species OYST is less than the median for the species BARN. First quartiles (Q1) for both data sets are approximately same, but medians and third quartiles are not same. Also, minimum observations for both data sets are same, however there is a difference in the maximum for the both data sets.

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The probability that at least one WST111 student is in time for the 7h30 lecture on a Tuesday morning is 0.9961.

1. First, let's find the probability that a randomly selected student is on time for the lecture. Since the probability that a student is late is 0.25, the probability that a student is on time is 1 - 0.25 = 0.75.

2. Now, we need to calculate the probability that all five randomly selected students are late for the lecture. Since punctuality is independent, we can simply multiply each student's probability of being late: 0.25×0.25×0.25×0.25× 0.25 = 0.0009765625.

3. Finally, we want to find the probability that at least one student is on time. To do this, we'll subtract the probability that all students are late from 1:

1 - 0.0009765625 = 0.9961.

So, the probability that at least one WST111 student is in time for the 7h30 lecture on a Tuesday morning is 0.9961.

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Answer:

Step-by-step explanation:

Solve for x.

4x - 9 = 2x + 5

4x - 2x = 5 + 9

2x = 14

x = 14 : 2

x = 7

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

check   (replace "x" with "7")

4 * 7 - 9 = 2 * 7 + 5                  (remember PEMDAS)

28 - 9 = 14 + 5

19 = 19

the answer is good

Answer:

hence the required value of x is 7.