Suppose a person is riding a plane 154ft above the ground and drops a box of supplies. The height of the box
is given by the formula:
h = -16+2 + 154
After how many seconds does the box hit the ground? (round to the nearest hundredth of a second)

Answer:

1.4

Step-by-step explanation:

Probability of selecting a red counter :

Number of red counter / total number of counters

5 / (5 + 3) = 5 /8 = 0.625

Probability of selecting a blue counter = 0.375

Let number of red counters :

0, 1 or 2

0 red counters :

Expected value = x * p(x)

0 + (0.625*1) + (0.390625 * 2)

0 + 0.625 + 0.78125 = 1.40625

Expected number of red counters = 1.4

Answer:

Point form -

[tex](2-\frac{5}{2})[/tex]

Equation form -

[tex]x=2, y=-\frac{5}{2}[/tex]

Step-by-step explanation:

if you have any questions feel free to ask

Since the mean and variance of y are the same as those of a normal random variable, y is a normal random variable.

The given moment generating function is Mx(t) = e^(t(4 + 5t + 17t^2/2)), where t is the moment. The moment generating function of a normal random variable is given by Mx(t) = e^(μt + σ^2t^2/2).Comparing the two, we get:μ = 4σ^2 = 17/2We can now compute the first and second moments of y, using the moment generating function: My(t) = e^(t(34 + 5t)) × e^(t(4 + 5t + 17t^2/2))= e^(34t + 9t^2 + 17t^3/2 + 4t + 5t^2) = e^(34t + 14t^2 + 17t^3/2)So,μy = My(0) = 1 and σy^2 = My''(0) - My'(0)^2= (17/2) - 4 = 9/2

Since the mean and variance of y are the same as those of a normal random variable, y is a normal random variable.

This completes the proof using the moment generating function uniqueness theorem.

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The moment generating function (MGF) of a random variable is a mathematical function that uniquely defines the probability distribution of that variable. Therefore, if the MGF of a random variable is the same as that of a known distribution, the random variable follows that distribution.

This is known as the uniqueness theorem of moment generating functions.

Using the given moment generating function of a random variable X (Max+b)(+).etMy.lat), we need to prove that X follows a normal distribution. Here, Max+b = a, My = m, and  = s (standard deviation).

Using the MGF, we can find the moments of the distribution. Differentiating the MGF 'n' times gives the nth moment about zero. We can use this to find the mean and variance of the distribution.

Therefore, the mean of the distribution is the first derivative of the MGF at t=0,

and the variance is the second derivative of the MGF at t=0.

Using facts (i) and (ii), we can write the MGF of X as:

(Mx(t)) = [(b + mt) + a(s^2 + m^2)/2] / [(s^2/2) + (t^2/2)]

Taking the first derivative of Mx(t) and substituting t=0, we get:

[tex]E(X) = [(b + m*0) + a(s^2 + m^2)/2] / [(s^2/2) + (0^2/2)]E(X) = (b + am) / (s^2/2) + 0E(X) = (2(b + am))/s^2[/tex]

This gives us the mean of the distribution as

(2(b + am))/s^2

Taking the second derivative of Mx(t) and substituting t=0, we get:

[tex]Var(X) = [(b + m*0) + a(s^2 + m^2)/2] / [(s^2/2) + (0^2/2)]Var(X) = [a(s^2 + m^2)/2] / (s^2/2) + 0Var(X) = a(s^2 + m^2)/s^2 - 1[/tex]

We know that the MGF of a normal distribution with mean m and variance s^2 is given by:

[tex](Mn(t)) = e^(mt + s^2t^2/2)[/tex]

Comparing this with the given MGF of X, we see that they are equal.

Therefore, X follows a normal distribution with mean (2(b + am))/s^2 and variance a(s^2 + m^2)/s^2 - 1.

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Step-by-step explanation:

3 coffees + 2 muffins = $7

therefore,

[tex]3x + 2y = 7[/tex]

2 coffees + 4 muffins = $8

therefore,

[tex]2x + 4y = 8[/tex]

Answer:

i think in every coast we're adding two shillings in the first sentence the total was 5 and we paid $7 and in the second sentence the total was 8 dollars meaning we added to shillings.

Step-by-step explanation:

in my school we are taughtthat we shouldn't mix goats with sheeps so we this the same way we can't mix x and y so in my opinion I guess you can say x + y =to that amount of money at the end you paid if it's 8 dollars or 7 dollars. we can say x represents the coffees and y represents the muffins that's how I get it I hope you want to be angry with me. for example 3 + 2 equals to 7 .3 + 2 is 5 so we'll carry five to the other side of equals where we will make it be minus so we'll say 7 - 5 where will get 2

To find the vector function that represents the curve of intersection between the paraboloid z = 2x^2y^2 and the parabolic cylinder y = 3x^2, we can parameterize the curve using a parameter t. The vector function r(t) will consist of x(t), y(t), and z(t), where each component is expressed in terms of t.

First, we need to find the relationship between x and y by setting the equation of the parabolic cylinder equal to the y-coordinate of the paraboloid. Substituting y = 3x^2 into z = 2x^2y^2, we get z = 2x^2(3x^2)^2 = 18x^6.

Now, we can express x and z in terms of t. Let's set x(t) = t, which allows us to write z(t) = 18t^6. The y-component can be obtained by substituting x(t) into the equation of the parabolic cylinder: y(t) = 3(t^2).

Finally, we can combine the x(t), y(t), and z(t) components to form the vector function r(t) = (x(t), y(t), z(t)). In this case, r(t) = (t, 3t^2, 18t^6) represents the curve of intersection between the two surfaces.

Note that this vector function parameterizes the curve and allows us to describe various points on the curve by plugging in different values of t.

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The probability represents the combination consists only of even numbers as per given condition is equal to 0.020.

Total numbers in combination lock = 38

Numbers from 0 to 37.

To find the probability that the combination consists only of even numbers,

Determine the total number of combinations that can be formed using only even numbers

And divide it by the total number of possible combinations.

Total number of even numbers in the lock

= 19 (since there are 19 even numbers from 0 to 37)

Calculate the total number of combinations using only even numbers,

Use the concept of combinations (nCr).

Since there are 19 even numbers to choose from,

Choose 4 numbers without repetition, the number of combinations is,

Number of combinations

= ¹⁹C₄

= 19! / (4!(19-4)!)

= (19 × 18 × 17 × 16) / (4 × 3 × 2 × 1)

= 3876

Now, calculate the total number of possible combinations without any restrictions.

Since we have 38 numbers to choose from,

and  choose 4 numbers without repetition, the number of combinations is,

Number of total combinations

= ³⁸C₄

= 38! / (4!(38-4)!)

= (38 × 37 × 36 × 35) / (4 × 3 × 2 × 1)

= 194,580

Finally, find the probability by dividing the number of combinations using only even numbers by the total number of combinations,

Probability

= Number of combinations using only even numbers / Total number of combinations

= 3876 / 194580

≈ 0.0199

≈ 0.020 Rounded to three decimal places.

Therefore, the probability that the combination consists only of even numbers is approximately 0.020.

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a)The force on electron B doubles the force acting on electron A

The correct answer is (b)

b)The path radius for electron A is half that of electron B .

The correct option is (a)

Moving charges are sources of the magnetic field and also recipients of the magnetic interaction. Stationary charges interact with the electric field but are not influenced by the magnetic field. The magnetic force acting on a moving charge is proportional to the value of the charge, to its velocity, and the absolute value of the magnetic field. The absolute value of the force also depends on the relative direction of the magnetic field and the velocity.

The magnetic force acting on the electron equals:

[tex]F_A =ev_A(B).....(1)\\\\F_B =ev_B(B).....(2)[/tex]

proportional to the vector product between the velocity and the magnetic field. The magnetic field and the velocity vectors are perpendicular, consequently, the absolute value of the vector product reduces to the product of the absolute values,

[tex]|F_A| =e|v_A||B|\\\\|F_B| =e|v_B||B|[/tex]

The only difference between the forces strives in the value of the velocities. The velocity of electron B is twice that of electron A . Therefore, the force on electron B doubles the force acting on electron A

The correct answer is (b)

b) To analyze the orbit of the electrons due to the magnetic force let us use Newton's second law. The magnetic force is always perpendicular to the velocity not changing its absolute value but only its direction. The acceleration resulting from the force is centripetal. In scalar form

[tex]F_A =ev_A(B)=ma_c_A\\\\F_B =ev_B(B)=ma_c_B[/tex]

Substituting the centripetal acceleration in terms of the velocities and the radiuses,

[tex]ev_AB=m\frac{v^2_A}{R_A} \\\\ev_BB=m\frac{v^2_B}{R_B}[/tex]

Solving for the radius,

[tex]R_A=\frac{mv_A}{eB} \\\\R_B=\frac{mv_B}{eB}[/tex]

The orbital radius is directly proportional to the velocity of the electron. As a result the path radius for electron A is half that of electron B .

The correct option is (a)

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The calculated value of the standardized score for this sample average is 2

From the question, we have the following parameters that can be used in our computation:

Population mean = 2

Population standard deviation = 0.6

Sample size = 36

Sample mean = 2.2

The standardized score for this sample average can be calculated using

z = (Score - Sample mean)/(Sample standard deviation)

Where

Standard deviation = Population standard deviation/√n

So, we have

Standard deviation = 0.6/√36

Evaluate

Standard deviation = 0.1

So, we have

z = (2.2 - 2)/0.1

Evaluate

z = 2

Hence, the standardized score for this sample average is 2

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Answer:

1 1/16

Step-by-step explanation:

you you have to add 7/8 + 3/16 to get the answer of one whole and 1/16

A 98% confidence interval for the variance of all student's grades is [9.41, 31.41].

The degrees of freedom (df) are (n - 1) = (20 - 1) = 19.

A confidence interval is given by the formula: $[{\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f},\ { \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}]$ where ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}$ is the lower percentile of the chi-square distribution with (n - 1) degrees of freedom, ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}$ is the upper percentile of the chi-square distribution with (n - 1) degrees of freedom, and s² is the sample variance.

Lower percentile (L.P) = ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}$

Upper percentile (U.P) = ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}$

Now, α = 0.02, n = 20, s² = 16, and df = 19.

Lower percentile:L.P = ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f} = { \chi }_{\frac{0.02}{2},19}^{2}\frac{16}{19} = 9.410$

Upper percentile:U.P = ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f} = { \chi }_{1-\frac{0.02}{2},19}^{2}\frac{16}{19} = 31.410$

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The 98% confidence interval for the variance of all student's grades is (10.12, 30.29). Hence, option A is the correct answer.

Given data are: The sample size, n is 20, the mean of the sample is 72 and the variance of the sample, s² is 16.

Option A is the correct answer.

Step-by-step explanation: Given, Sample size, n = 20

Mean of the sample, [tex]\bar x = 72[/tex]

Variance of the sample, [tex]s^{2} =16[/tex].

We have to find a 98% confidence interval for the variance of all student's grades.

Assumption: Grades are distributed normally.

So, the formula for the confidence interval for the variance is: [tex]((n-1)s^{2} / \chi^{2} \alpha /2, (n-1)s^{2} / \chi^{2} 1-\alpha /2)[/tex]

Where, [tex]\alpha = 1-0.98[/tex]

= 0.02

n-1 = 19

Degrees of freedom [tex]\chi^{2}\alpha /2[/tex] and [tex]\chi^{2}²1-\alpha /2[/tex] are the critical values of chi-square distribution with (n-1) degrees of freedom.

From the chi-square distribution table, we can find the values of [tex]\chi^{2}\alpha /2[/tex] and [tex]\chi^{2}²1-\alpha /2[/tex] such that the area between these values is 0.98.

Here, [tex]\alpha = 0.02[/tex] (as 98% confidence interval)

[tex]\alpha /2 = 0.02/2[/tex]

= 0.01

Using the above values, we get,

[tex]\chi^{2}0.01/2,19 = 8.907[/tex] and

[tex]\chi^{2}1-0.01/2,19 = 32.852[/tex]

Now, using the formula of confidence interval for the variance, we get,

[tex]((n-1)s^{2} / \chi^{2} \alpha /2, (n-1)s^{2} / \chi^{2} 1-\alpha /2)=((20-1) \times 16 / 32.852, (20-1) \times 16 / 8.907)[/tex]

We get, (10.12, 30.29).

Therefore, the 98% confidence interval for the variance of all student's grades is (10.12, 30.29). Hence, option A is the correct answer.

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Given: The mass of a species of mouse commonly found in houses is normally distributed with a mean of 20.9 grams with a standard deviation of 0.2 grams.

a) The probability that a randomly chosen mouse has a mass of less than 20.74 grams is 0.2119.

b) The probability that a randomly chosen mouse has a mass of more than 21.14 grams is 0.1151.

c) The proportion of mice have a mass between 20.8 and 21.04 grams is 0.4495.

d) 10% of all mice have a mass of less than 21.15632 grams.

(a) The probability that a randomly chosen mouse has a mass of less than 20.74 grams.

Z = (X - μ)/σ, where X = 20.74, μ = 20.9, σ = 0.2.

Z = (20.74 - 20.9)/0.2Z

= -0.8P(X < 20.74)

= P(Z < -0.8)

Using standard normal distribution table, P(Z < -0.8) = 0.2119.

Thus, the required probability is 0.2119.

(b) The probability that a randomly chosen mouse has a mass of more than 21.14 grams.

Z = (X - μ)/σ, where X = 21.14, μ = 20.9, σ = 0.2.

Z = (21.14 - 20.9)/0.2

Z= 1.2

P(X > 21.14) = P(Z > 1.2)

Using standard normal distribution table, P(Z > 1.2) = 0.1151.

Thus, the required probability is 0.1151.

(c) To find the proportion of mice have a mass between 20.8 and 21.04 grams.

Z_1 = (X1 - μ)/σ

= (20.8 - 20.9)/0.2

= -0.5

Z_2 = (X2 - μ)/σ

= (21.04 - 20.9)/0.2

= 0.7

P(20.8 < X < 21.04) = P(-0.5 < Z < 0.7)

= P(Z < 0.7) - P(Z < -0.5)

Using standard normal distribution table,

P(Z < 0.7) = 0.7580

P(Z < -0.5) = 0.3085

Therefore, P(-0.5 < Z < 0.7) = P(Z < 0.7) - P(Z < -0.5)

= 0.7580 - 0.3085

= 0.4495

Thus, the required probability is 0.4495.

(d) 10% of all mice have a mass of less than grams.

Let the mass be X.

We have to find X such that P(X < k) = 0.1

Z = (X - μ)/σ, where μ = 20.9, σ = 0.2.

P(X < k) = P(Z < (k - μ)/σ)

0.1 = P(Z < (k - 20.9)/0.2)

0.1 = P(Z < 5k - 104.5)

Using standard normal distribution table, we get

5k - 104.5 = -1.2816k

k= 21.15632

Thus, 10% of all mice have a mass of less than 21.15632 grams.

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Answer:

Step-by-step explanation:

se gana 10000 y 2

To find the confidence interval for p1 - p2 at a given level of confidence 95%, with x1 = 354, n1 = 545, x2 = 406, n2 = 596, we need to first calculate the point estimate for the difference in proportions:

$$\hat{p_1} = \frac{x_1} {n_1} = \frac {354}{545} \approx. 0.6495$$$$\hat{p_2} = \frac{x_2} {n_2} = \frac {406}{596} \approx. 0.6822$$

Therefore, the point estimate of the difference in proportions is: $$\hat{p_1} - \hat{p_2} = 0.6495 - 0.6822 \approx. -0.0327$$. Now, we can use the formula for the confidence interval for the difference in proportions:

$$\text{Confidence interval} = (\hat{p_1} - \hat{p_2}) \pm z_{\alpha/2} \sqrt{\frac{\hat{p_1}(1 - \hat{p_1})}{n_1} + \frac{\hat{p_2}(1 - \hat{p_2})}{n_2}}$$where z_{\alpha/2} is the z-score for the level of confidence 95% (or 0.95), which is approximately 1.96

Using this information, the confidence interval for p1 - p2 at a 95% level of confidence is: $$(0.6495 - 0.6822) \pm 1.96 \sqrt {\frac {0.6495(1 - 0.6495)}{545} + \frac {0.6822(1 - 0.6822)}{596}}$$$$\approx. -0.0327 \pm 0.0472$$

Therefore, we can conclude that the researchers are 95% confident the difference between the two population proportions, P1 - P2, is between -0.080 and -0.004 (using ascending order and rounding to three decimal places as needed).

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Answer:

eight.

Step-by-step explanation:

2(6) ÷ 3

12÷3

4+4

8

There are 3150 ways the Kickers soccer team can field a 2/4/4 team.

To determine the number of ways the Kickers soccer team can field a 2/4/4 team, we need to consider the different positions and the number of players available in each position.

For a 2/4/4 team, we have:

2 strikers

4 midfielders

4 full-backs

2 goalies

To calculate the number of ways, we can multiply the number of choices for each position:

Number of ways = Number of choices for strikers [tex]\times[/tex] Number of choices for midfielders [tex]\times[/tex] Number of choices for full-backs [tex]\times[/tex] Number of choices for goalies.

For the strikers, there are 4 available players, and we need to choose 2 of them.

This can be calculated using the combination formula, denoted as C(n, r):

Number of choices for strikers = C(4, 2) = 4! / (2! [tex]\times[/tex] (4-2)!) = 6

For the midfielders, there are 6 available players, and we need to choose 4 of them:

Number of choices for midfielders = C(6, 4) = 6! / (4! [tex]\times[/tex] (6-4)!) = 15

Similarly, for the full-backs, there are 7 available players, and we need to choose 4 of them:

Number of choices for full-backs = C(7, 4) = 7! / (4! [tex]\times[/tex] (7-4)!) = 35

Finally, for the goalies, there are 2 available players, and we need to choose 2 of them:

Number of choices for goalies = C(2, 2) = 1

Now, we can multiply all the choices together:

Number of ways [tex]= 6 \times15 \times 35 \times 1 = 3150[/tex]

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A

see attached for explanation

I hope it helps

A performance task is a learning activity that asks students to perform to demonstrate their knowledge, understanding and proficiency.

Your information is incomplete. Therefore, an overview will be given. A performance task measure whether a student can apply his or her knowledge to make sense of a new phenomenon.

It should be noted that performance tasks yield a tangible product and performance that serve as evidence of learning.

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??? The answer is not able to be found for the fact that there are 3 numbers all coming from x.

Answer:

Below.

Step-by-step explanation:

The formula for a GS is:

a1, a1r,  a1 r^2, a1r^3 .....

Examples are:

1. First  term a1 =2 and common ratio r = 3, the first 4 terms are:

2,  2*3,  2*3^2,  2*3^3

= 2, 6, 18, 54

2. a1 = 1 , r = 0.5:

1, 0.5,  0.25,  0.125  .....

3. a1 = -4,  r = -2:-

-4,  8, -16,  32  ...

4.  a1 = 0.1,  r = -3:

0.1,  -0.3,  0.9, -2.7 .....

5. a1 = 2, r = 1/3:

2,  2/3,  2/9,  2/27 .....

The correct option is:d.9/2 pie units cubbed. The given region is obtained by rotating the curve y = x2 – 3 about the x-axis, which can be represented as:y = x² - 3

We have to rotate this curve around the y-axis. The required volume can be obtained by integrating the area of the circular cross-sections formed after rotating the curve.The radius of each circular cross-section at any value of x is equal to the perpendicular distance between the curve and the y-axis, which is x.

Thus, the area of each circular cross-section can be represented as:A = π(x²)The limits of integration can be found by equating y = x² - 3 to 0, which is:0 = x² - 3x = ± √3

Thus, the volume of the solid obtained by rotating the region bounded by y = x² - 3 and the x-axis about the y-axis is given by:V = ∫-√3^√3 π(x²) dxV = π ∫-√3^√3 (x²) dx.

By using the formula for integration, we get:V = π [(x³)/3] -√3^√3V = π [(√3³ - (-√3)³)/3]V = π (18/3)V = 6π.Therefore, the volume of the solid obtained by rotating the region bounded by y = x² - 3 and the x-axis about the y-axis is 6π units cubed. Hence, the correct option is:d.9/2 [tex]\pi[/tex]units cubbed.

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Answer:

Triangle JKLM ~ Triangle STUV

(a) The possible remainders when 12 + 16 + 20 is divided by 11 are 8 and 19.

(b) For every integer n, the expression 121 + n^2 + 16n + 20 is always divisible by 11.

In the first statement, we are asked to prove that a - c divides ab + cd if and only if a - c divides ad + bc. To prove this, we can use the property of divisibility. If a - c divides ab + cd, then there exists an integer k such that ab + cd = (a - c)k. Similarly, if a - c divides ad + bc, there exists an integer m such that ad + bc = (a - c)m. By rearranging the terms, we can express k and m in terms of a, b, c, and d. By substituting these expressions into the equation ab + cd = (a - c)k, we can show that ab + cd = (a - c)(ad + bc)m. Thus, proving the equivalence.

In the second statement, we are asked about the possible remainders when 12 + 16 + 20 is divided by 11. To find the remainder, we can calculate the sum as 48 and then divide it by 11. The quotient is 4 with a remainder of 4. Hence, 12 + 16 + 20 leaves a remainder of 4 when divided by 11.

To prove that 121 + n^2 + 16n + 20 is divisible by 11 for every integer n, we can use algebraic manipulation. By factoring the expression as (n + 4)(n + 4) + 16(n + 4), we can rewrite it as (n + 4)^2 + 16(n + 4). Now, we notice that both terms have a common factor of (n + 4). Factoring it out, we get (n + 4)(n + 4 + 16). This simplifies to (n + 4)(n + 20). Since both factors contain n + 4, we can conclude that the expression is divisible by 11 for any integer n.

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The combination of pies she can make are:

one cherry pie, one peach pie and one apple pie

A bar graph is defined as a diagram in which the numerical values of variables are represented by the height or length of lines or rectangles of equal width.

We are given the following parameters:

Number of apples to make one apple pie = 6

Number of peaches to make one peach pie = 9

Number of cherries to make one cherry pie = 32

From the bar graph, she has:

26 apples

12 peaches

34 cherries

Since she wants to make 3 pies, then she can make:

one cherry pie, one peach pie and one apple pie

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Answer: 60

Step-by-step explanation:

Answer:24 or 4

Step-by-step explanation:

2x2=4 4x6=24 24 or 4

Answer:

g(f(x)) = 2x - 6

f(g(x)) = 2x - 4

Step-by-step explanation:

g(f(x)) = 2x - 8 + 2

g(f(x)) = 2x - 6

f(g(x)) = 2(x + 2) - 8

f(g(x)) = 2x - 4

Answer: g ( 2 x − 8 ) = 2 x− 6

Step-by-step explanation: MAKE ME BRAINLIEST!!!!!

Answer:

x=0

Step-by-step explanation:

Answer:

Short answer: Empty set ∅

if x is less than -1 then it can't be greater than 1

The solution to the given initial value problem is u(x,t) = [e−(z−¹)² + e¯(x+t)²]. [infinity] < x

Given Differential equation is J²u(x, t) Ət² = dx²I.

C is u(x,0) = e-x² and ut (x,0) = 0

We need to find the solution to the given initial value problem. The general solution to the given Differential equation is

u(x,t) = f1(x − t) + f2(x + t)

where f1 and f2 are arbitrary functions. We need to find the particular solution that satisfies the given initial conditions. Given intial condition,

u(x,0) = e-x² …………(1)

ut (x,0) = 0…………(2)

Let us find the value of f1(x) and f2(x) using the given intial condition,

From equation (1)u(x,0) = f1(x) + f2(x) = e-x²

Now, Differentiating f1(x) + f2(x) = e-x² w.r.t x, we getf1'(x) + f2'(x) = -2x ………..(3)

From equation (2)

ut (x,0) = f1'(x) + f2'(x) = 0

On solving the above two equations, we get

f1'(x) = x, f1(x) = ½x² + C1

and

f2'(x) = -x, f2(x) = ½x² + C2

Now, the particular solution is

u(x,t) = ½(x − t)² + C1 + ½(x + t)² + C2

u(x,t) = (x² + t² + C1 + C2)………….(4)

Using the initial condition u(x,0) = e-x² in equation (4)

e-x² = x² + C1 + C2

Now, we know that for large value of x, the term e-x² becomes negligible. So, C1 + C2 = 0 ⇒ C2 = -C1

Substituting C2 = -C1 in equation (4), we get

u(x,t) = [e−(z−¹)² + e¯(x+t)²]. [infinity] < x

The solution to the given initial value problem is u(x,t) = [e−(z−¹)² + e¯(x+t)²]. [infinity] < x.

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