What are the dangers you should look out for before using the AED?.

Answer: 3 and 1

Step-by-step explanation: 3 is a prime number, which means its only factors are the number itself and 1. The number 3 is not divisible by anything else. You can only multiply 3 and 1.

Integral value is finite since the given integral converges when the function is [tex]\int\limits^\infty_\infty {3xe^{-x^{2} } } \, dx[/tex].

Given that,

Analyze the integral to see if it is convergent or divergent. convergent integrals should be evaluated.

[tex]\int\limits^\infty_\infty {3xe^{-x^{2} } } \, dx[/tex]

We have to simplify the equation.

Finding the area of the curve's undersurface is the process of integration. To do this, draw as many little rectangles as necessary, then add up their areas.

We know that,

I= [tex]\int\limits^\infty_\infty {3xe^{-x^{2} } } \, dx[/tex]

Let p = x²then xdx= dp/2

Then

I= [tex]\int\limits^\infty_\infty {3/2e^{-p} } \, dp[/tex]

I= 3/2([tex]e^{-p}[/tex])infinity to minus infinity

I= 3/2 ([tex]e^{-x^{2} }[/tex])infinity to minus infinity

I=0

Therefore, integral value is finite since the given integral converges when the function is [tex]\int\limits^\infty_\infty {3xe^{-x^{2} } } \, dx[/tex].

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The foot and orientation of the normal vector of this plane whose ray would cross (1,1,1) is (0,1,1) and 2.

Given:

we have three points:

Q(1,0,0),

R(0,1,0),

and S(0,0,1)

when the plane passes through Q,R, and S, then the vectors

QR = (-1,1,0) and QS = (-1,0,1)

lie in the plane. Thus the cross-product

n = I i  j  k  I

      -1 1 0

      -1 0 1 I

= i + j + k is normal to the plane.

If r=⟨x,y,z⟩ determines an arbitrary point P in the plane, the vector

v = QP = ⟨x-1,y−0,z−0⟩

ray = (1,1,1)

v = QP = (1-1, 1-0, 1-0)

= (0,1,1)

lies in the plane and so is perpendicular to n. In this case,

n⋅v = 1(0)+1(1)+1(1)

= 0+1+1

= 2

Hence the foot and orientation f the normal vector of this plane is (0,1,1) and 2.

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Answer:

Yes, the measurements 3cm, 4cm and 5cm are possible to create a triangle!

Step-by-step explanation:

The rate at which the volume of the box of height 10 inches is changing when the width is 8 inches and the length is 6 inches is 440 cubic inches per second

The volume of a rectangular box is the product of the length, the width and the height of the box.

The height of the rectangular box, h = 10 inches

The rate of change of the length of the box is; [tex]\frac{dl}{dt}[/tex] = 2 in/sec

The rate at which the width is decreasing is; [tex]\frac{dw}{dt}[/tex] = 4 in/sec

When the length is 8 inches and the width is 6 inches, we get;

[tex]\dfrac{dV}{dt} = \dfrac{d}{dt} \left(l\cdot w \cdot h\right)[/tex]

[tex]\dfrac{dV}{dt} =h\times \left(l\cdot \dfrac{dw}{dt} + w \cdot \dfrac{dl}{dt} \right)[/tex]

When l = 8, and w = 6, we get;

[tex]\dfrac{dV}{dt} =10\times \left(8\times 4 + 6 \times 2 \right)=440[/tex]

The rate at which the volume of the rectangular box is changing in cubic inches per second is therefore;

[tex]\dfrac{dV}{dt} =440 \ in^3/sec[/tex]

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Using the Pythagoras threoem,

the possible value of k are 6 or 1 in right angled triangle.

We have given that,

There is a triangle say, ∆ABC with vertices A(1,2) ,B(k , 2) and C(6,7).

and angle, B = 90°

that is this is a right angled triangle with angle ABC is 90° .

we have to find out the value of k .

Using Pythagoras threoem,

AB² + BC² = AC²---(*)

where, AB ,BC and AC are sides of ∆ABC.

using distance formula, we find length of sides of ∆ABC . let coordinates of A (a₁,b₁) and B(a₂,b₂)

AB = √(a₂ - a₁)² + (b₂-b₁)²

put a₁ = 1 , b₁ = 2 , a₂ = k , b₂ = 2

AB² = (k - 1)² + (0) = (k-1)²

similarly we can easily calculate the length of other sides of ∆ABC.

BC² = (6-k)² + 25

AC²= (5)² + (5)² = 50

putting all sides value in above formula,

50 = (k-1)² + (6-k)²+ 25

=> 25 = k² + 1 - 2k + 36 + k² - 12k

=> 2k² - 14k + 12 = 0

=> k²- 7k +6 = 0

this is a quadratic equation we solve it and find out value of k

using , roots formula

k = { -(-7) +- √(-7)^2 - 4(1)(6)}/2×1

=> k = either( 7 + √25) /2 or (7- √25)/2

=> k = either 6 or 1

Hence , the possible value of k are 6 or 1

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The conclusion of step 4 and step 5 are not valid . Therefore, this is an invalid proof that all isosceles triangles.

Each step is valid, but the conclusion is not. At the end of step 4, there are a series of isometric maps that lead A to D and C to F, as claimed.. After the final rotation in Step 5, the sequence of isometries takes B to E and C to F —but here’s the hitch—there is no reason why it should still take A to D, in fact it can’t take A to D unless the rotation angle is zero.

If the rotation angle in Step 5 is zero, then the isometries take [A,B,C] to [D,E,F] and the two triangles are indeed similar. But this requires the dilation in Step 2 to have made A′B′ congruent to DE , in other words, it requires assuming the triangles are similar before performing any of of the transformations.

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Complete Question:

This is an invalid proof that all isosceles triangles are similar. Explain which step is invalid and why. 1. Draw 2 isosceles triangles ABC and DEF where AC = BC and DF = EF. 2. Dilate triangle ABC to a new triangle A'B'C' using center C and scale factor DF/AC so that A'C= BC - DF = EF. 3. Translate by directed line segment CF to take A'B'C' to a new triangle A" B" F. Since translation preserves distance, A"F = A'C=DF and B" F = BC = EF. 4. Since A"F=DF, we can rotate using center F to take A" to D. 5. Since B" F = EF, we can rotate using center F to take B" to E. 6. We have now established a sequence of dilations, translations, and rotations that takes A to D, B to E, and C to F, so the triangles are similar.

The given pair of events are independent when your first coin flip results in heads. and your second coin flip results in heads.

Events that occur independently of any other events are referred to as independent events. if we flip a coin in the air and get the head, we will then flip the coin again, but this time we will get the tail. The occurrence of both events is unrelated in either instance. Tossing the coin the first time does not affect the outcome of the second toss.

Note that the events are not mutually exclusive as it is possible that both the first and the second toss are headed.

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It gradually decreases as alcohol is metabolized. The function C(t)=0.135 t e^{-2.802 t}models the mean BAC measured in g/mL.

The maximum average BAC during 3 hours is 0.0001358 g/mL.

f(t) = α t e−βt --(1)

Let's rewrite this in a simple form:

f(t)= α eˡⁿ ᵗ e⁻βt = αe^(ln t −βt)

Since e^x is strictly increasing and it will be maximized exactly when its argument is maximized, so we can maximize instead:

g(t)=ln t −βt

differentiating with respect to t , and g'(t) = 0

g′(t)=1/t − β = 0

=> t =1/β

we have given a function

C(t)=0.135 t e⁻²·⁸⁰²ᵗ

if we compare it with (1) we get

β = 2.802, 0.135 = α

For it's maximized we need to check the second order condition, and that of g will differentiate again , g′′(t)= −1/t² < 0

We have to compute the derivative of C(t):

C′(t) = 0.135 t⋅(−2.802)e⁻²·⁸⁰²ᵗ + 1.35e⁻²·⁸⁰²ᵗ

For optimum at t₀ if C′(t₀)=0 and C′′(t₀)≠0. Here, we have

C′(t₀) = 0.135t₀⋅(−2.802)e⁻²·⁸⁰²ᵗ₀+ 0.135e⁻²·⁸⁰²ᵗ₀ =e⁻²·⁸⁰²ᵗ₀(−0.135* 2.802t₀+ 0.135)=0

It is clear that e⁻²·⁸⁰²ᵗ₀ not equal to zero for all t₀∈R, so that

=> −0.135* 2.802t₀+0.135=0

=> t₀ = 1/2.802 ≈0.36

let us consider t is in hours, so that it makes t₀ =0.36h≈21.41min. This is the only optimum and one should verify it is indeed a maximum, i.e. C′′(t₀)<0.

Now, easily compute the maximum average BAC, which is C(t₀)=C(0.36) = 0.135 (0.36)e⁻²·⁸⁰²⁽⁰·³⁶⁾

= 0.0486(0.3678) = 0.01787508

Hence, the maximum average BAC, is 0.017 g/dL.

Maximum average BAC during the first 3 hours,

t = 3 , C(t)=C(3) = 0.135 (3)e⁻²·⁸⁰²⁽³⁾ = 0.0001358 g/mL

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The answer is all real numbers greater than or equal to -3, as domain and range.

What is the domain and range of function?

A function's domain and range are its components. A function's range is its potential output; its domain is the set of all conceivable input values. Domain, Function, and Range. If a function f: A B exists that maps every element in A to an element in B, then A is the domain and B is the co-domain. The image of an element 'a' under a relation R is provided by 'b,' where (a,b) R. The set of photographs makes up the function's range.

The domain of a relation or a function is the set of all possible values for an input in that relation/function.

The range of a relation or function is the set of all possible values that can be the result of a relation or function.

From the graph, nothing can be seen on the left side of x = - 3 and the graph is continuous and going in the direction of infinity on the right side of the x axis.

Hence all real numbers greater than or equal to -3

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The component form of the force vector required to move the object of 300 pounds is (272, 127)

Suppose we have a vector v  with angle θ to the positive x-axis, then the horizontal and vertical components of the vector are given as:

vx = v cos θ

vy = v sin θ

In the given problem, the vector is a force vector of 300 pounds.

w = 300

θ = 25⁰

Hence,

horizontal component = wx = w cos 25⁰

                                         = 300 . cos 25⁰ = 272

vertical component = wy = w sin 25⁰

                                         = 300 . sin 25⁰ = 127

Hence, the component form of the given force vector is (272, 127)

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Answer:

Its D on edge

Step-by-step explanation:

Answer:

Step-by-step explanation:

solving by elimination means to get rid of one of the variables,  pick either one,  X or Y  and use the other equation to make it zero.

3x - y = 5

4x + y = 9

+______  add the two equations, column for column,

7x + 0y = 14

now solve for x,  that is, get x by itself

7x / 7 = 14/7

x = 2

Now that you've solved for x,  plug that x into either equation.

3(2) - y = 5

6-y = 5

6-5 = y +5-5

1 = y

Now you have both :)

At the null hypothesis, it is tested if there is not enough evidence that the means are difference, that is, if the subtraction is of zero, hence:

[tex]H_0: \mu_d = 0[/tex]

At the alternative hypothesis, it is tested if there is enough evidence that the means are difference, that is, if the subtraction is different of zero, hence:

[tex]H_1: \mu_d \neq 0[/tex]

We have a two-tailed test, as we are testing if the mean is different of a value, with 5 + 5 - 2 = 8 df and a significance level of 0.05, hence the critical value is of:

|t| = 2.306.

For each sample, the mean and the standard error are given as follows:

For the distribution of differences, the mean and the standard error are given as follows:

The test statistic is given by the division of the sample mean by the standard error, hence:

t = -11.8/17.19

t = -0.69.

The absolute value of the test statistic is less than the critical value for the two-tailed test, hence:

The data is given by the image at the end of the answer.

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Answer: Answer is 3241 gm or 3.2 kg approx.

Step-by-step explanation:

Weight of the brick( in kg) =22 kg

Weight of the brick in gm= 22*1000 gm      (1 kg = 1000gm)

                                          = 22000 gm

Weight of bricks removed = 18759 gm

Weight of bricks left in the pile= 22000-18759gm

                                                   =3241 gm or 3.2 kg approx

Hence, the weight of the bricks left is 3.2 kg approx.

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The point estimate of the population mean number of units sold per month is given by:

90.

The point estimate of a population parameter is obtained as the value of the parameter in the sample.

The parameter that we want to obtain the estimate in this problem is:

The mean.

The sample in this problem is given by:

94, 85, 85, 94, 92.

(number of units sold each month).

The sample mean is given by the sum of all observations in the sample divided by the number of observations, hence it is obtained as follows:

Mean = (94 + 85 + 85 + 94 + 92)/5 = 90.

For the fraction, we have that:

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The correct answer is amount of dilute solution is 4 gallons and amount of concentrated solution is 6 gallons.

Let the dilute amount of solution be x and the amount of salt in the solution be 0.13x

Then the concentrated amount be 10-x  and the amount of salt in the solution be 0.33x (x-10).

Solving the equation

⇒ 0.13x + 0 .33 (10 - x ) = 2.5

⇒ 0.13x + 3.3 - 0.33x  = 2.5

⇒ -0.2 x = 2.5 - 3.3

⇒ -0.2x = - 0.8

⇒ x = 4

Hence, the amount of dilute solution is 4 gallons and amount of concentrated solution is 6 gallons.

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Answer: 20 & 37

What we know:

Process:

10*2=20

20 is a product of 10 because (10*2=20)

37 is not a product of 10 because there are no whole numbers that 10 can be multiplied with to make 37

In the given triangle, the length of missing side is 15.

A triangle is a geometric figure with three edges, three angles and three vertices. It is a basic figure in geometry.

The sum of the angles of a triangle is always 180°

In the given triangle,

Side BD = 5, CE = 3 and AE = 9

Let the side AD is x,

Since, line BC and DE and parallel to each other,

Therefore, by side splitter theorem,

A line that is parallel to one side of a triangle and crosses the other two sides equally divides the side in proportional ratio.

Implies that,

AD/BD = AE/EC

x/5 = 9/3

x/5 = 3

x = 15

The value of side AD is 15.

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The slope intercept form of the given line equation is y=-5/6 x+1.

The given equation of a line is 5x+6y=6.

The slope intercept form of a straight line is one of the most common forms used to represent the equation of a line. The slope intercept formula can be used to find the equation of a line when given the slope of the straight line and the y-intercept.

The standard form of the slope intercept form is y=mx+c.

Here, the equation can be written in slope intercept form as

5x+6y=6

6y=6-5x

y=-5/6 x+1

Therefore, the slope intercept form of the given line equation is y=-5/6 x+1.

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Answer: 28

Step-by-step explanation: a true false question has 2 questions per question if it either true or false so if there is 14 questions with all true and false it would be 14 x 2 = 28

The correct option is B is correct that is by setting access to private on the class fields using the correct notation.

Given that,

Data hiding refers to the protection of sensitive data from outside code for an object. This is achieved using Options

A) Making use of the class methods' private access specifier.

B) By setting access to private on the class fields using the correct notation.

C) Using the class methods' public access specifier.

D) Using the class definition's private access specifier.

Whichever choice is accurate must be determined.

We know that,

The private modifier limits access to a class's methods or properties from any code that is outside the class, but the public access modifier permits a code from inside or outside the class to access the class's methods and properties.

Therefore, The correct option is B is correct that is by setting access to private on the class fields using the correct notation.

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Answer:

Please see the attachment below.

Step-by-step explanation:

Please give the brainliest, really appreciated.

Using simple Units Measurement System,

Dressmaker cut total 40 pieces of length 18 inch from 15 of given rolls of ribbon.

We have given that,

total length of roll of ribbon = 6 feet

the length of ribbon cut by dressmaker = 18 inch

As we know , 1 inch = 2.5 cm --(1)

then 1 feet = 30cm --(2)

from (1) and (2) , we get

1 feet = 12 inch

total length of roll ribbon in inch = 6×12

= 72 inches

we have total number of ribbons = 5

so, total number of inches = 5 ×72 = 360 inches

As per question given that dressmaker need to cut 18 inch piece for making dress .

So, total number of 18 inch pieces Dressmaker has is equal to ratio of total length of ribbon / length of ribbon dressmaker needed

= 720/18

= 40

Hence, total 40 pieces of length 18 inch dressmaker have.

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The COUNTIF function to count the number of students who have been elected to offices in student organizations in Cell Q8 is ‘=COUNTIF(M2:M31, "Elected")’.

The COUNTIF function is a premade function in Excel, which counts cells as specified. It is one of the statistical functions which is used to count the number of cells that meet a criterion. In the given situation, Lael wants to count the number of students who have been elected to offices in student organizations from the data available in the excel spreadsheet. The COUNTIF function is used to determine the total by using the formula: ‘=COUNTIF(M2:M31, "Elected")’ in the Cell Q8.

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The true statement is option (a) If the sample size is bigger, the interval is narrower.

Given:

for a given confidence level and population or sample standard deviation the following is true in the interval estimation of the population mean.

In a confidence interval, to find the standard error we divide the standard deviation by the sample size. When the sample size is large the value of standard error decreases. When the value of standard error decreases the width of the confidence interval also decreases. Therefore we can say If the sample size is bigger, the interval is narrower.

that is,

error α standard deviation / sample size

Here when sample size is large the error decreases because both are inversely proportional when error decreases the standard deviation or confidence intervals decrease because both are directly proportional.'

Therefore The true statement is option(a) If the sample size is bigger, the interval is narrower.

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Full question:

For a given confidence level and population standard deviation, which of the following is true in the interval estimation of the population mean?

(a) If the sample size is bigger, the interval is narrower.

(b) If the sample size is smaller, the interval is narrower.

(c) If the population size is bigger, the interval is narrower.

(d) If the population size is smaller, the interval is narrower.

By using the concept of relation, it can be proved that

If R is a Reflexive relation, then [tex]R^{-1}[/tex] is a Reflexive relation

If R is a Symmetric relation, then [tex]R^{-1}[/tex] is a Symmetric relation

If R is a Transitive relation, then [tex]R^{-1}[/tex] is a Transitive relation

What is a relation?

Relation on a set [tex]X_1, X_2,....,X_n[/tex] is the subset of the cartesian product

[tex]X_1 \times X_2 \times..... \times X_n[/tex]

A)

Let R be a reflexive relation .

Let (x, x) ∈ R

Then (x, x) ∈ [tex]R^{-1}[/tex]

So [tex]R^{-1}[/tex] is reflexive

B)

Let R be a symmetric relation such that [tex]xR^{-1}y[/tex] holds
so yRx holds

Since R is symmetric, xRy holds

[tex]yR^{-1}x[/tex] hold

[tex]R^{-1}[/tex] is symmetric

C)

Let R be a transitive relation such that [tex]xR^{-1} y[/tex] and [tex]yR^{-1}z[/tex] holds

Then yRx and zRy holds

Since R is transitive

zRx holds

[tex]xR^{-1}z[/tex] holds

[tex]R^{-1}[/tex] is transitive

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Complete Question

Let R Be A Relation On A Set A. Prove The Following: A) If R Is Reflexive, Then [tex]R^{-1}[/tex] Is Reflexive. B) If R Is Symmetric, Then [tex]R^{-1}[/tex]Is Symmetric. C) If R Is Transitive, Then [tex]R^{-1}[/tex] Is Transitive.

The speed at 60 miles to 2 dp if Karina didn't stop, will be 38.5 mph.

Speed is defined as the rate of change of position of a subject in any direction. We have the basic formula to find the speed given by:

Speed = [tex]\frac{Distance}{Time}[/tex]
s = [tex]\frac{d}{t}[/tex]

To find the average speed for the last 60 miles, we need to know the time spent by Karina during the final 60 miles.

For the initial time during 70 miles, we follow the below calculations:

Initial Speed = [tex]s_{i}[/tex] = 50mph
Initial Distance = [tex]d_{i}[/tex] = 70 miles
Initial time for 70 miles = [tex]t_{i}[/tex]

[tex]s_{i}[/tex] = [tex]\frac{d_{i} }{t_{i} }[/tex]

[tex]t_{i}[/tex] = [tex]\frac{d_{i} }{s_{i}}[/tex]

[tex]t_{i}[/tex] = [tex]\frac{70}{50}[/tex] = 1.4 hr

For the Total time = t,

Total Speed = s = 44 mph.

Total Distance = d = [tex]d_{i} + d_{f}[/tex] = 70 + 60 = 130 miles

Total time = [tex]\frac{d}{t}[/tex] = [tex]\frac{130}{44}[/tex] = 2.96 hr

For the Final time during 70 miles = [tex]t_{f}[/tex]

t = [tex]t_{i} + t_{f}[/tex]
[tex]t_{f}[/tex] = t - [tex]t_{i}[/tex]
[tex]t_{f}[/tex] = 2.96 - 1.4 = 1.56 hr

Hence, the average speed for the final 60 miles can be calculated as:

Final Speed = [tex]s_{f}[/tex] = ?
Final Distance = [tex]d_{f}[/tex] = 60
Final Time = [tex]t_{f}[/tex] = 1.56 hr.

Average Speed = [tex]s_{a}[/tex] =  [tex]\frac{d_{f}}{t_{f}}[/tex] =  [tex]\frac{70}{1.56}[/tex] = 38.50 mph.

So, the Final average speed for the last 60 miles Karina traveled is 38.50 mph

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Answer:

91.1 cm

Step-by-step explanation:

the formula to find circumferences is pi times diameter

29pi

91.1

Hopes this helps please mark brainliest

The school will purchase 80 chairs and 10 tables.

A system having more than two equations is known as a system of equations.

Given that, the school can spend $5200 and the cost of one table is $200 and the cost of one chair is $40.

Let x represents the number of chairs purchased and y represents the number of tables purchased.

The total cost of purchasing x chairs and y tables is:

Total cost = 40x + 200y

Since the budget of the school is $5200 only, substitute "Total cost =5200" into the above equation,

5200 = 40x + 200y

Since each table requires 8 chairs, it follows:

x = 8y

Hence, the system of equations is:

5200 = 40x + 200y      (1)

x = 8y                            (2)

To solve the system of equations, substitute x = 8y into equation (1):

5200 = 40(8y) + 200y

5200 = 320y + 200y

5200 = 520y

y = 10

Substitute y = 10 into equation (2):

x = 8(10)

x = 80

Hence, the school will purchase 80 chairs and 10 tables.

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The E(T) is 4 when A and B, two players, alternately toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on).

Given that,

A and B, two players, alternately toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on). The heads and tails are recorded in order. The game is over and the winner is the person who throws the tail if there is a head followed by a tail (HT sub-sequence).

We have to find what is the anticipated game length. Let the game continue for T number of coin tosses. Find E (T).

We know that,

Here,

E(T )=first time sequence HT occurs

E(T) = P(first H)× P(expected number till 1st T |first H)+ P(first T)× P(expected number till 1st HT |first T)

E(T) =0.5×(1+2)+0.5×(1+E(T))

0.5×E(T)=0.5×4

E(T) =4

Therefore, The E(T) is 4 when A and B, two players, alternately toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B and so on).

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