7. suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. determine a differential equation for the number of people x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between the number of students who have the flu and the number of students who have not yet been exposed to it. (Use k > 0 for the constant of proportionality and x for x(t).)

The region of the concealed district will be 226.08 square centimeters.

It is the nearby bend of an equidistant point drawn from the middle. The sweep of a circle is the distance between the middle and the boundary.

Let d be the diameter of the circle. Then the area of the circle will be written as,

A = (π/4)d² square units

A ring-formed district's internal measurement is 14 cm and its external breadth is 22 cm. Then the region of the concealed district is given as,

A = (π / 4) (22² - 14²)

A = (3.14 / 4) (484 - 196)

A = 0.785 x 288

A = 226.08 square cm

The region of the concealed district will be 226.08 square centimeters.

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The system of equations are solved

a) The train will reach at 2:00 PM if it leaves at 11:25 AM

b) The train will reach at 21:20 PM if it leaves at 18:45 PM

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

Substituting the values in the equation , we get

A train journey from London to Leed takes 2h 35min

So , the total journey time is 155 minutes

a)

The time when the train reaches Leeds when it leaves at 11:25 AM is given by the equation A = 11:25 AM + 155 minutes

On simplifying the equation , we get

The train will reach at 2:00 PM if it leaves at 11:25 AM

b)

The time when the train reaches Leeds when it leaves at 18:45 PM is given by the equation A = 18:45 PM + 155 minutes

On simplifying the equation , we get

The train will reach at 21:20 PM if it leaves at 18:45 PM

Hence , the equations are solved

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The number of calories burned per minute be 23.33333.

Equations simultaneously, or a system of equations Several equations in algebra must be solved concurrently (i.e., the solution must satisfy all the equations in the system). There must be an equal number of equations and unknowns for a system to have a singular solution.

In order to locate the point where the lines intersect when the equations are graphed, systems of equations must be solved. The (x,y) ordered pair of this intersection point is regarded as the system's solution.

Let j be the number of calories burned by Shelly while running and c be the number of calories she burns while cycling.

The system of equations be

45j + 30c = 1350

30j + 45c = 1350

45j + 30(10)=1350

j = 23.33333

Therefore, the value j be 23.33333.

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The relative frequency of landing on heads is 0.4, then the correct option is D.

When we have an experiment with some outcomes, and we perform the experiment N times, and in K of these N times we get a particular outcome, then the relative frequency for that outcome is K/N

In this case the coin is flipped 35 times and it lands on heasd 14 times, then the relative frequency of landing on heads is:

R = 14/35 = 0.4

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Answer:

The final height of the tower is 114 feet 8 inches.

To solve this problem, we need to convert the additional 8 inches into feet. 8 inches is equal to 0.67 feet, so the new height of the tower is equal to 114 feet + 0.67 feet which is equal to 114 feet 8 inches.

The missing value is given as follows:

P(X = 28) = 0.31.

The mean and the standard deviation are given as follows:

The sum of the probabilities of all the outcomes is of one, hence the missing value is obtained as follows:

0.14 + 0.18 + 0.21 + P(X = 28) + 0.16 = 1

0.69 + P(X = 28) = 1

P(X = 28) = 0.31.

The mean is given by the sum of all outcomes multiplied by their respective probabilities, hence:

E(X) = 25 x 0.14 + 26 x 0.18 + 27 x 0.21 + 28 x 0.31 + 29 x 0.16

E(X) = 27.17.

The standard deviation is given by the square root of the sum of the difference squared between each observation and the mean, multiplied by their respective probabilities, hence:

S(X) = sqrt((25-27.17)² x 0.14 + (26-27.17)² x 0.18 + (27-27.17)² x 0.21 + (28-27.17)² x 0.31 + (29-27.17)² x 0.16)

S(X) = 1.289.

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Answer:

Step-by-step explanation:

c

Using relational algebra:

a. flight ⋈ airport ⋈ ρ destination  airport

b. flight ⋈ airport

c. passenger ⋈ airport

d. airline ⋈ airport ⋈ flight

e. airline ⋈ airport ⋈ flight

f. airline ⋈ airport ⋈ flight ⋈ ρ destination airport

g. flight ⋈ passenger ⋈ booking

QUESTION (a):

- using EQUI join (⋈) join relations "Flight, Airport and Airport as destination"

- and using select Operation (σ), relational operators (=, >) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "Flight.fID, Airport.code, destination.code, Flight.fare" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π flight . fid, airport . code, destination . code , flight.fare   σ flight . airport = airport . code

    AND flight . destination = destination . code  (flight ⋈ airport ⋈ ρ destination  airport)

QUESTION (b):

- using EQUI join (⋈) join relations "Flight, Airport"

- and using select Operation (σ), relational operators (=, >=, <=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airport . city, airport . code, flight . departure" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airport . city, airport . code, flight . departure    σ flight . airport = airport . code

     AND (4 <= flight . departure AND flight . departure <= 10)  (flight ⋈ airport)

QUESTION (c):

- using EQUI join (⋈) join relations "passenger, Airport"

- and using select Operation (σ), relational operators (<>) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "passenger . name, passenger . hometown" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π passenger . name, passenger . hometown    σ passenger . hometown <> airport . city  (passenger ⋈ airport)

QUESTION (d):

- using EQUI join (⋈) join relations "airline, airport, flight"

- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

    AND flight . airport = airport . code

    AND airport . city = "Toronto"  (airline ⋈ airport ⋈ flight)

QUESTION (e):

- using EQUI join (⋈) join relations "airline, airport, flight"

- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

     AND flight . airport = airport . code

    AND airport . city <> "Toronto"  (airline ⋈ airport ⋈ flight)

QUESTION (f):

- using EQUI join (⋈) join relations "airline, airport, flight and airport as destination"

- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

    AND flight . airport = airport . code

    AND flight . destination = destination . code

    AND airport . city = "Toronto"

    AND destination . city = "vancouver"  (airline ⋈ airport ⋈ flight ⋈ ρ destination airport)

QUESTION (g):

- using EQUI join (⋈) join relations "flight, passenger, Booking"

- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "passenger . name, airport, destination" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π passenger . name, airport, destination    σ passenger . pid = booking . pid

    AND booking . fid = flight . fid

    AND destination . city = "vancouver"  (flight ⋈ passenger ⋈ booking)

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Question correction:

See on the attached image.

The probability is P(k) = (q^(k-1)) * p for k>=1, and P(0) = q. The expected number of times the loop will be executed is 1/p.The expected number of times that the repeat loop will be executed is 1/p.

To find the probability that the loop will be executed k times, we can consider the probability of the event that B is false k-1 times followed by B being true. This probability is q^(k-1) * p.

The event of the loop not being executed at all corresponds to B being true in the first trial, which has a probability of q. Therefore, the probability that the loop will be executed k times is P(k) = (q^(k-1)) * p for k>=1, and P(0) = q.

The expected number of times the loop will be executed is the sum of the probabilities of executing the loop k times, weighted by k, i.e., E = Sum(kP(k)) for k>=1, and E = 0 if P(0) = q.

By using the expression for P(k), we can simplify this to E = Sum(kq^(k-1)*p) for k>=1, and E = 0 if P(0) = q. By applying the formula for the sum of a geometric series, we get E = 1/p.

For the "repeat S until B" loop, the expected number of times that the loop will be executed is the expected number of trials in a Bernoulli process until the first success, where the success probability is p. By using the formula for the expected value of a geometric distribution, we get E = 1/p.

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The part of the expression x(5 + y) that goes to the parentheses is 5 + y.

The correct option is 4.

One mathematical expression makes up a term. It might be a single variable (a letter), a single number (positive or negative), or a number of variables multiplied but never added or subtracted. Variables in certain words have a number in front of them. A coefficient is a number used before a phrase.

Given:
A phrase: “x times the quantity 5 plus y.”

5 plus y 5 + y

x times the quantity 5 plus y = x(5 + y)

The complete expression is,

x(5 + y).

Therefore, 5 + y is the required expression.

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Answer: 1/9

Step-by-step explanation:

2/2=1

18/2=9

The approximate measure of angle G.The correct answer is a. 41.4 degrees.

The measure of angle G in triangle EFG can be calculated using the Pythagorean Theorem. The Pythagorean Theorem states that the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. In right triangle EFG, the two legs are EF and EG and the hypotenuse is FG. This can be expressed mathematically as [tex]8^2 + 10^2 = 12^2.[/tex] Simplifying the expression, the equation becomes 64 + 100 = 144. Solving this equation yields 64 = 144, which is true. To calculate the measure of angle G, we will use the inverse tangent function, which is written as [tex]tan^-1[/tex]. In this function, the inverse tangent of the ratio of the opposite side to the adjacent side is equal to the angle. This can be expressed mathematically as [tex]tan^-1 (8/10)[/tex] = G. Using a calculator, the inverse tangent of 8/10 is approximately 41.4 degrees. Therefore, the correct answer is a. 41.4 degrees.

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Answer:6

Step-by-step explanation:

ƒ(4)=4+2=6

The two sets of lengths that do not form a triangle are the options B and C.

For a triangle with side lengths x, y, and z we know the triangular inequality, it says that the sum of any two sides must be larger than the other side, so we can write 3 inequalites:

x + y > z

x + z > y

z + y > x

So if for one of the given sets, one of these inequalities is false, then the set does not form a triangle.

For the second set:

12 yd, 25 yd, 13 yd

The inequality:

12 yd + 13yd > 25yd

25 yd> 25 yd

is false, so that set does not form a triangle.

And the last set:

5 ft, 9 ft, 16 ft​

The inequality:

5ft + 9ft > 16ft

14ft > 16 ft

Is also false,

So B and C are the correct options.

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The price of burger is  $3.75 and the fries cost $1.25

On Monday, Jack bought 2 burgers and 3 fries for $11.25

On Tuesday he bought 7 burgers and 5 fries for $32.50

Let a represent the cost of the burger

Let b represent the cost of the fries

2a + 3b= 11.25..........equation 1

7a + 5b= 32.50..........equation 2

Solve by elimination method
Multiply equation 1 by 7 and multiply equation 2 by 2

14a + 21b= 78.75

14a + 10b= 65

11b= 13.75

b= 13.75/11

b = 1.25

Substitute 1.25 for b in equation 2

7a + 5(1.25)= 32.50

7a + 6.25= 32.50

7a= 32.50-6.25

7a= 26.25

a= 26.25/7

a= 3.75

Hence the price of burger is $3.75 and the price of fries is $1.25

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Therefore, the number that satisfies all the given conditions is 60,649.

In mathematics, an equation is a statement that asserts the equality of two expressions, typically separated by an equals sign ("="). The expressions on either side of the equals sign are called the left-hand side and the right-hand side of the equation, respectively. The purpose of an equation is to describe a relationship between two or more variables or quantities, such as x + 3 = 7 or y = 2x - 5. Equations can be used to solve problems and answer questions in various fields of study, such as algebra, geometry, physics, chemistry, and engineering. Solving an equation typically involves finding the value or values of the variable(s) that make the equation true. Some equations may have a unique solution, while others may have multiple solutions or no solutions at all. The study of equations and their properties is a fundamental topic in mathematics.

Here,

Let's break down the clues given in the problem and use them to find the unknown number:

6 ten thousands: The number must start with 6.

2 fewer thousands than ten thousands: The number of thousands is 2 less than the number of ten thousands. Since there are 6 ten thousands, there are 4 thousands.

Same number of hundreds as ten thousands: The number of hundreds is the same as the number of ten thousands, which is 6.

1 fewer ten than ten thousands: The number of tens is 1 less than the number of ten thousands, which is 6-1=5.

5 more ones than thousands: The number of ones is 5 more than the number of thousands, which is 4+5=9.

Putting all of these clues together, we get the number: 60,649

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The integral for the volume of the solid is obtained by rotating the region bounded by the given curves about the specified axis is  V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]

When we rotate a thin vertical strip, about the y-axis.

We get a cylindrical shell  with an inner of radius an x and an outer of radius x + dx

The height of the cylinder shell is tan (x) - x

The volume of the cylindrical shell is

     dV = π [tex](Outer Radius)^{2} (Height)[/tex] - π[tex](Inner Radius)^{2} (Height)[/tex]

     dV = π [tex](x + dx )^{2} (tan(x) - x)[/tex] - π [tex](x )^{2} (tan(x) - x)[/tex]

dV = π[tex](x^{2} + 2 xdx + (dx)^{2} ) (tan(x) - x)[/tex] - π [tex](x)^{2} (tan(x) - x)[/tex]

assume [tex]dx^{2}[/tex]≈ 0

            dV = π[tex](x^{2} + 2xdx + 0 - x^{2} ) (tan(x) - x)\\[/tex]

                    dV = 2πx (tan(x) - x) dx

                   V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]

Therefore, the integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis is  

V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex] .

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The correct question is:

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=tanx,y=x,x=π/3; about the y-axis

Answer:

90c^2 -30c - 420

Step-by-step explanation:

-3(-3c+7)5(4+2c)

(9c - 21) (20 + 10c)

180c + 90c^2 - 420 - 210c

90c^2 -30c - 420

Since the p-value is less than .05, we reject the null hypothesis that the means of the three assembly methods are equal.

Therefore, we can conclude that there is a significant difference in the means of the three assembly methods.

Source Variation | Squares' Sum | Degrees of Freedom | Mean Square | F

Treatments 4560     2        2280     9.87

Error         6240     27      231.11

Total         10800   29

Using Alpha = .05 to test for any significant difference in the means for the three assembly methods.

The value of the test statistic is 9.87

The p-value is: less than .01

Conclusion not all means of the three assembly methods are equal.

Complete Question:

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,800; SSTR = 4560.

Set up the ANOVA table for this problem (to 2 decimals, if necessary).

The probability of getting 3 heads in 4 coin tosses, given that we get at least 2 heads, is 4/11 or 0.364.

To solve this problem, we can use the conditional probability formula. Let A be the event of getting 3 heads in 4 coin tosses, and let B be the event of getting at least 2 heads in 4 coin tosses. Then we want to find P(A|B), the probability of getting 3 heads in 4 coin tosses given that we get at least 2 heads.

By the definition of conditional probability, we have:

P(A|B) = P(A and B) / P(B)

To find P(B), the probability of getting at least 2 heads in 4 coin tosses, we can use the complement rule and find the probability of getting 0 or 1 heads:

P(B) = 1 - P(0 heads) - P(1 head)

To find P(0 heads), the probability of getting 0 heads in 4 coin tosses, we use the binomial probability formula:

P(0 heads) = (4 choose 0) * (0.5)^0 * (1-0.5)^(4-0) = 1/16

Similarly, we can find P(1 head):

P(1 head) = (4 choose 1) * (0.5)^1 * (1-0.5)^(4-1) = 4/16

So,

P(B) = 1 - P(0 heads) - P(1 head) = 11/16

To find P(A and B), the probability of getting 3 heads in 4 coin tosses and getting at least 2 heads, we can use the binomial probability formula again:

P(A and B) = (4 choose 3) * (0.5)^3 * (1-0.5)^(4-3) = 4/16

Therefore,

P(A|B) = P(A and B) / P(B) = (4/16) / (11/16) = 4/11

So the probability of getting 3 heads in 4 coin tosses, given that we get at least 2 heads, is 4/11 or approximately 0.364.

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The proof of LMNO is a rhombus is shown below.

A parallelogram is a particular instance of a rhombus. The opposing sides and angles in a rhombus are parallel and equal. A rhombus also has equal-length sides on each side, and its diagonals meet at right angles to form its shape. The rhombus is also referred to as a diamond or rhombus.

Given:

|LO|=|MN| and |LM|=|ON|

Since Opposite sides of a parallelogram are equal.

Now, LN⊥OM

So, ∠LPO = ∠NPO = 90° ( by definition of perpendicular lines)

LPO ≅ ∠NPO (by definition of congruent angles)

|LP|=|PN| (diagonals of a parallelogram bisect each other)

Thus, LMNO is a rhombus

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Answer:

♣: ✔ All right angles are congruent.

♦: ✔ reflexive property

♠: ✔ Opposite sides of a parallelogram are congruent.

a) To compare the overall percentage of late deliveries for each delivery service, we need to calculate the total number of late packages and total number of deliveries for each service, and then divide the number of late packages by the total number of deliveries, and multiply by 100 to get the percentage.

For Pack Rats, the total number of deliveries is 400 + 100 = 500, and the total number of late packages is 12 + 16 = 28. Therefore, the percentage of late deliveries for Pack Rats is:

percentage of late deliveries for Pack Rats = (28/500) x 100 = 5.6%

percentage of late deliveries for Boxes R Us = (30/500) x 100 = 6%

Therefore, based on these calculations, Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us.

b) While Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us, it's important to note that this decision should not be based solely on this one comparison. It's possible that there are other factors that the company needs to consider, such as the cost of each delivery service or the quality of customer service provided by each company.

Furthermore, the sample size in this trial period may not be large enough to draw a definitive conclusion about the reliability of each service. Therefore, while Pack Rats may be a good choice based on the available data, the company should consider other factors and conduct further research before making a final decision.

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The Mean Absolute Deviation (MAD) for actual sales is 300 and for forecast is 600.

Mean Absolute Deviation evaluates the absolute difference between each data point to its mean. Mean Absolute Deviation can be calculated using formula:

MAD = ∑|x - x bar|

                n

where:

x = data point

x bar = data mean

n = number of data

Based on the given data, we know that:

Mean Actual sales = (1,000 + 1,600 + 2,000 + 1,800) / 4

Mean actual sales = 1,600

MAD = |1,000 - 1,600| + |1,600 - 1,600| + | 2,000 - 1,600| + |1,800 - 1,600|

                                                4

MAD = (600 + 0 + 400 + 200) / 4

MAD = 300

Mean Forecast = (600 + 2,500 + 1,500 + 2,000) / 4

Mean forecast = 1,650

MAD Frc = |600 - 1,650| + |2,500 - 1,650| + |1,500 - 1,650| + |2,000 - 1,650|

                                                                4

MAD Forecast = (1,050 + 850 + 150 + 350) / 4

MAD Forecast = 600

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Answer: 60 = 1/√3.

Step-by-step explanation:

The cotangent of 60 degrees is equal to the x-coordinate of the point of intersection divided by the y-coordinate of the point of intersection. In this case, the x-coordinate is 0.5 and the y-coordinate is √3/2. Therefore, cot 60 = 0.5 / √3/2 = 1/√3.

So, cot 60 = 1/√3.

The trapezoid ABCD not isosceles because AB is not congruent to DC.

It is a polygon that has four sides. The sum of the internal angle is 360 degrees. In a trapezoid, one pair of opposite sides are parallel.

Quadrilateral ABCD has vertices A(-3,4), B(2,5), C(3,3), and D(-1,0).

The diagram is given below.

From the diagram, the line segment AD and BC are parallel to each other.

The length AB is given as,

AB² = (2 + 3)² + (4 - 5)²

AB = 5.1 units

The length CD is given as,

CD² = (3 + 1)² + (3 - 0)²

CD= 5 units

The trapezoid ABCD not isosceles because AB is not congruent to DC.

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No, it is not safe to stop the addition of terms when their magnitude falls below s, even if the desired absolute error is less than e.

This is because the magnitude of the terms in the series may not decrease monotonically, and there may be large fluctuations in the magnitudes of the terms.
Therefore, it is necessary to use convergence tests, such as the ratio test or the root test, to determine if the series converges absolutely.

For the series ∑ (0.99)^n, we can use the ratio test to check for absolute convergence:

lim (n → ∞) |(0.99)^(n+1)/(0.99)^n| = 0.99 < 1

Since the limit is less than 1, the series converges absolutely. However, we cannot simply stop adding terms when their magnitude falls below a certain value s, as the magnitude of the terms in the series may not decrease monotonically.

Instead, we need to use the convergence test to determine the number of terms required to achieve the desired absolute error e.

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Answer: The frequency (f) of a wave is related to its wavelength (λ) and velocity (v) by the equation:

f = v/λ

Given the wavelength range of 100 m to 1000 m for the medium radio wave band, we can calculate the frequency range as follows:

For the lower wavelength of 100 m:

f = v/λ = 299.8 × 10^6 m/s / 100 m = 2.998 × 10^6 Hz

For the higher wavelength of 1000 m:

f = v/λ = 299.8 × 10^6 m/s / 1000 m = 299.8 × 10^3 Hz

Therefore, the frequency range for the medium radio wave band is approximately 2.998 × 10^6 Hz to 299.8 × 10^3 Hz.

Step-by-step explanation:

Answer: Given the formula:

V = 6pirt + 4pir^2

To solve for t, we'll isolate t by rearranging the equation.

First, subtract 4pir^2 from both sides:

V - 4pir^2 = 6pirt

Next, divide both sides by 6pi:

(V - 4pir^2)/6pi = t

So, t = (V - 4pir^2)/6pi.

This gives us the value of t in terms of V and the radius of the cylinder, r.

Step-by-step explanation: