Suppose that neighborhood soccer players are selling raffle tickets for $500 worth of groceries at a local store, and you bought a $1 ticket for yourself and one for your mother. The children eventually sold 1000 tickets. What is the probability that you will win first place while your mother wins second place?

[tex]HTT,HTH,HHH,TTT,THT,THH,TTH[/tex]

Final answer:

The probability that the second character will die given that the first character dies is 53.33%. This is known as conditional probability.

Explanation:

To find the probability that the second character will die given that the first character dies, we use the concept of conditional probability.

The formula for conditional probability is P(B|A) = P(A and B) / P(A), where P(B|A) is the probability of event B occurring given that event A has occurred, P(A and B) will be the probability of both events A and B occurring, and P(A) is the probability of event A occurring.

In this scenario, event A is the first character dying, and event B is the second character dying. The student has already stated there is a 70% chance that the first character will survive, which means there is a 30% (100% - 70%) chance that the first character will die.

They've also stated a 16% chance that both characters will die. Applying the formula gives us P(B|A) = P(A and B) / P(A) = 0.16 / 0.30 = 0.5333, or 53.33%.

Therefore, the probability that the second character will die given that the first character dies is 53.33%. This kind of probability is called conditional probability.

Answer: 0.33

Step-by-step explanation:

Let,

In this question we are using the Bayes' theorem,

where,

P(E1) = P(E2) = P(E3) = [tex]\frac{1}{3}[/tex]

As there is an equal probability assign for choosing a coin.

Given that,

it comes up heads

so, let A be the event that heads occurs

then,

P(A/E1) = 1

P(A/E2) = 0

P(A/E3) =  [tex]\frac{1}{2}[/tex]

Now, we have to calculate the probability that the opposite side of coin is tails.

that is,

P(E3/A) = ?

∴ P(E3/A) = [tex]\frac{P(E3)P(A/E3)}{P(E1)P(A/E1) + P(E2)P(A/E2) + P(E3)P(A/E3) }[/tex]

= [tex]\frac{(1/3)(1/2)}{(1/3)(1) + 0 + (1/2)(1/3)}[/tex]

= [tex]\frac{1}{6}[/tex] × [tex]\frac{6}{3}[/tex]

= [tex]\frac{1}{3}[/tex]

= 0.3333 ⇒ probability that the opposite face is tails.

Given a double-headed coin, a double-tailed coin, and a regular coin, the probability that the opposite face is tails after tossing a head is 33.33%, assuming we picked one coin randomly and tossed it to see a head.

The student is asking about a problem involving conditional probability, with the specific condition that one of the sides that came up is a head. We are given three coins: a double-headed coin, a double-tailed coin, and a regular coin. The aim is to calculate the probability that the opposite face is tails given that the tossed coin shows heads.

First, we need to consider the total number of heads that can come up when choosing any coin. This yields two heads from the double-headed coin, and one head from the regular coin, resulting in three possible heads. However, only the regular coin has a tail on the opposite side.

Consequently, the probability that the opposite face is tails given that a head has been tossed is 1 out of 3, or 33.33%.

Answer:

if x= -1 then its is NOT in the domain of h.

Step-by-step explanation:

Domain is the set of values for which the function is defined.

we are given the function

[tex]h(x) =\frac{x+1}{x^2 + 2x + 1}[/tex]

Solving the denominator by factorization

[tex]h(x) =\frac{x+1}{x^2 + x+x + 1}\\h(x) =\frac{x+1}{x(x+1)+1(x + 1)}\\h(x) =\frac{x+1}{(x+1)(x+1)}\\h(x) =\frac{1}{(x+1)}[/tex]

So, if x = -1 then its is NOT in the domain of h.

Divide both sides by [tex]x^2-1[/tex] to get a linear ODE,

[tex]\dfrac{\mathrm dy}{\mathrm dx}+\dfrac2{x^2-1}y=\dfrac{x+1}{x-1}[/tex]

In order for this operation to be valid in the first place, we require that [tex]x\neq\pm1[/tex] (since that would make [tex]\dfrac1{x^2-1}[/tex] undefined, which we don't want to happen). Then we are forcing any solution to the ODE to exist on any of the three intervals, [tex](-\infty,-1)[/tex], [tex](-1, 1)[/tex], or [tex](1,\infty)[/tex], and either the first or third of these can be chosen as the largest interval.

In case you also need to solve the ODE: Multiply both sides by [tex]\dfrac{1-x}{1+x}[/tex], so that

[tex]\dfrac{1-x}{1+x}\dfrac{\mathrm dy}{\mathrm dx}-\dfrac2{(1+x)^2}y=-1[/tex]

Then the left side can be condensed as the derivative of a product, since

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}\right]=-\dfrac2{(1+x)^2}[/tex]

and we have

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}y\right]=-1[/tex]

Integrate both sides:

[tex]\displaystyle\int\frac{\mathrm d}{\mathrm dx}\left[\frac{1-x}{1+x}y\right]\,\mathrm dx=-\int\mathrm dx[/tex]

[tex]\dfrac{1-x}{1+x}y=-x+C[/tex]

[tex]\implies\boxed{y=\dfrac{(-x+C)(1+x)}{1-x}}[/tex]

The largest interval over which the general solution is defined for the given differential equation is [-1, ∞).

Here's how:

Answer: [tex]H_0:\mu\geq33[/tex]

[tex]H_a:\mu<33[/tex]

Step-by-step explanation:

Let [tex]\mu[/tex] be the average number of hours a person drive without adding the product.

Given claim : An infomercial claims that a woman drove 33 hours without​ oil.

i.e. [tex]\mu\geq33[/tex]

It is known that the null hypothesis always contains equal sign and alternative hypothesis is just opposite of the null hypothesis.

Thus the null and alternative hypothesis for the given situation will be :-

[tex]H_0:\mu\geq33[/tex]

[tex]H_a:\mu<33[/tex]

Answer: 0.3679

Step-by-step explanation:

The formula for Poisson distribution  :-

[tex]P(x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}[/tex]

Let x be the number of breakdowns.

Given : The rate of breakdown per week :  0.5

Then , for 2 weeks period the number of breakdowns = [tex]\lambda=0.5\times2=1[/tex]

Then , the probability that there will be no breakdown on his car in the trip is given by :-

[tex]P(x)=\dfrac{e^{-1}1^{0}}{0!}=0.367879441171\approx0.3679[/tex]

Hence, the required probability : 0.3679

Answer:

The probability is 0.2356.

Step-by-step explanation:

Let X is the event of using the smartphone in meetings or classes,

Given,

The probability of using the smartphone in meetings or classes, p = 51 % = 0.51,

So, the probability of not using smartphone in meetings or classes, q = 1 - p = 1 - 0.51 = 0.49,

Thus, the probability that fewer than 5 of them use their smartphones in meetings or classes.

P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3)+P(X=4)

Since, binomial distribution formula is,

[tex]P(x) = ^nC_r p^x q^{n-x}[/tex]

Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Here, n = 11,

Hence,  the probability that fewer than 5 of them use their smartphones in meetings or classes

[tex]=^{11}C_0 (0.5)^0 0.49^{11}+^{11}C_1 (0.5)^1 0.49^{10}+^{11}C_2 (0.5)^2 0.49^{9}+^{11}C_3 (0.5)^3 0.49^{8}+^{11}C_4 (0.5)^4 0.49^{7} [/tex]

[tex]=(0.5)^0 0.49^{11}+11(0.5)0.49^{10} + 55(0.5)^2 0.49^{9}+165 (0.5)^3 0.49^{8} +330(0.5)^4 0.49^{7} [/tex]

[tex]=0.235596671797[/tex]

[tex]\approx 0.2356[/tex]

Answer:

The only bijection is f(x)=2x+1.

I took r to be a constant.

Step-by-step explanation:

Bijections are both onto and one-to-one.

Onto means every element of the codomain gets hit.  Here the codomain is the set of real numbers.  So you want every y to get hit.

One-to-one means you don't want any y to get hit more than once.

f(x)=2x+1 is a linear function.  It is diagonal line so every element of the codomain will get hit and hit only once so this function is onto and one-to-one.

f(x)=x^2+1 is a quadratic function.  It is parabola so not every element of our codomain will get not get hit and of those that do get hit they get hit more than once.  So this is neither onto or one-to-one.

f(x)=r^3 is a constant function.  It is a horizontal line so not every y will get hit so it isn't onto.  The same y is being hit multiple times so it isn't one-to-one.

f(x)=(x^2+1)/(r^2+2) is a quadratic. It is a parabola. Quadratic functions are not onto or one-to-one.

Answer:

72 months approx.

Step-by-step explanation:

Monthly deposit = m = $140

r = 3.8% or 0.038

Amount needed in the account = A = $11300

The formula will be :

[tex]11300=140(\frac{(1+0.038/12)-1}{0.038/12} )[/tex]

[tex]11300=140(\frac{(1+0.038/12)-1}{0.003166})[/tex]

[tex]11300=44219.83((1.003166)^{m}-1)[/tex]

[tex]0.2555=(1.003166)^{m}-1[/tex]

[tex]1.2555=(1.003166)^{m}[/tex]

m=log1.2555/log1.003166

m =71.98 ≈ 72

Hence, it will take 72 months approx.

Answer:

  a) at 16%: $10,936.47

  b) at 8%: $124,177.02

Step-by-step explanation:

At annual rate of return "r", the multiplier of Sarah's initial investment will be ...

  k = (1+r)^34

For r = 0.16, k ≈ 155.433166, and Sarah's investment needs to be ...

  $1.7·10^6/k ≈ $10,936.47

__

For r= 0.08, k ≈ 13.6901336, and Sarah's investment needs to be ...

  $1.7·10^6/k ≈ $124,177.02

Answer:

Yes

Step-by-step explanation:

We are given that a function [tex]\phi(x)=x^2-x^{-1}[/tex]

We have to find that given function is an explicit solution to the linear equation

[tex]\frac{d^2y}{dx^2}-\frac{2}{x^2}y=0[/tex]

If given function is an explicit solution of given linear equation then it satisfied the given differential equation

Differentiate w.r.t x

Then we get [tex]\phi'(x)=2x+x^{-2}[/tex]

Again differentiate w.r.t x

Then we get

[tex]\phi''(x)=2-\frac{2}{x^3}[/tex]

Substitute all values in the given differential equation

[tex]2-\frac{2}{x^3}-\frac{2}{x^2}(x^2-x^{-1})[/tex]

=[tex]2-\frac{2}{x^3}-2+\frac{2}{x^3}=0[/tex]

Hence, given function is an explicit solution of given differential equation.

Therefore, answer is yes.

Answer: 0.3439

Step-by-step explanation:

Total number of digits : 10

The number of digits except zero =9

The probability of selecting non-zero number :-

[tex]\dfrac{9}{10}=0.9[/tex]

Then, the probability that one such phone number , the last four digits do not include any zero:-

[tex]\text{P(no zero )}=(0.9)^4[/tex]

Then , the probability that for one such phone​ number, the last four digits include at least one 0:-

[tex]\text{P(at-least one zero )}=1-(0.9)^4=0.3439[/tex]

Hence, the probability that for one such phone​ number, the last four digits include at least one 0 is 0.3439 .

To find the probability of at least one 0 in a four-digit number, one can use the complement of the probability of no 0s occurring in any of the digits. The probability of at least one 0 is approximately 34.39%.

The student is inquiring about the probability of a certain event, specifically related to randomly generating telephone numbers. To find the probability that at least one digit is a 0 in a randomly generated four-digit number, we can use the complementary probability approach. The probability of not getting a 0 in one digit is 9/10 since there are 9 other possible digits (1-9). Hence, the probability of not getting a 0 in any of the four digits is (9/10)^4. Subtracting this from 1 will give the probability of having at least one 0 in the four-digit sequence:

Probability of at least one 0 = 1 - (Probability of no 0 in any digit)

Probability of at least one 0 = 1 - (9/10)^4

After calculating, we have:

Probability of at least one 0 = 1 - 0.6561 = 0.3439

Therefore, the probability of having at least one 0 in a randomly chosen telephone number with four digits is approximately 34.39%.

Answer:

A. number of decayed atoms = 73.197

Step-by-step explanation:

In order to find the answer we need to use the radioactive decay equation:

[tex]N(t)=N0*e^{kt}[/tex] where:

N0=initial radioactive atoms

t=time

k=radioactive decay constant

In our case, when t=0 we have 1,000,000 atoms, so:

[tex]1,000,000=N0*e^{k*0}[/tex]

[tex]1,000,000=N0[/tex]

Now we need to find 'k'. Using the provied information that after 365 days we have 973,635 radioactive atoms, we have:

[tex]973,635=1,000,000*e^{k*365}[/tex]

[tex]ln(973,635/1,000,000)/365=k[/tex]

[tex] -0.0000732=k[/tex]

A. atoms decayed in a day:

[tex]N(t)=1,000,000*e^{-0.0000732t}[/tex]

[tex]N(1)=1,000,000*e^{-0.0000732*1}[/tex]

[tex]N(1)= 999,926.803[/tex]

Number of atoms decayed in a day = 1,000,000 - 999,926.803 = 73.197

B. Because 'k' represents the probability of decay, then the probability that on a given day 51 radioactive atoms decayed is k=0.0000732.

Answer:

B. 200 doses

Step-by-step explanation:

Given,

1 dose is required for 100 mg,

Since, 1 mg = 0.001 g,

⇒ 100 mg = 0.1 g

⇒ 1 dost is required for 0.1 g,

Thus, the ratio of doses and quantity ( in gram ) is [tex]\frac{1}{0.1}=10[/tex]

Let x be the doses required for 20 grams,

So, the ratio of doses and quantity is [tex]\frac{x}{20}[/tex]

[tex]\implies \frac{x}{20}=10[/tex]

[tex]\implies x=200[/tex]

Hence, 200 doses can be obtained from 20 grams of the drug.

Option 'B' is correct.

To solve this question, we will follow these steps:
1. We need to ensure that we use the same units for both the drug amount and the dose. Since the drug amount is given in grams and the dose in milligrams, we will convert the drug amount from grams to milligrams.
2. We know that 1 gram is equivalent to 1000 milligrams.
3. Now, let's convert the drug amount from grams to milligrams:
  We have 20 grams of the drug, so the conversion to milligrams is:
  \(20 \text{ grams} \times \dfrac{1000 \text{ milligrams}}{1 \text{ gram}} = 20,000 \text{ milligrams}\)
4. Next, we will divide the total milligrams of the drug by the milligram dosage that is recommended per dose to find out how many doses we can get from the drug amount.
5. Given that each dose is 100 mg, we now divide the total drug amount in milligrams by the dose in milligrams:
  \(20,000 \text{ milligrams} \div 100 \text{ milligrams per dose} = 200 \text{ doses}\)
Therefore, from 20 grams of the drug, we can obtain 200 doses.
The correct answer is:
  B. 200 doses

A horizontal line is a line where all of the [tex]y[/tex] values are the same. In this case, [tex]\boxed{y=5}[/tex], so that is the equation.

A vertical line is where all of the [tex]x[/tex] values are the same.  Here, [tex]\boxed{x=7}[/tex], so that's the equation.

Answer:

see below

Step-by-step explanation:

A horizontal line has the same y value  and has a constant y value

y=5

A vertical line has the same x value  and has a constant x value

x=7

Answer:

8.82 years.

Step-by-step explanation:

Since, the monthly payment formula is,

[tex]P=\frac{PV(r)}{1-(1+r)^{-n}}[/tex]

Where, PV is the present value of the loan,

r is the rate per month,

n is number of months,

Here,

PV =  $ 25,000,

Annual rate = 5.2 % = 0.052 ⇒ Monthly rate, r = [tex]\frac{0.052}{12}[/tex]

( 1 year = 12 months )

P = $ 295,

By substituting the values,

[tex]295=\frac{25000(\frac{0.052}{12})}{1-(1+\frac{0.052}{12})^{-n}}[/tex]

By the graphing calculator,

We get,

[tex]n = 105.84[/tex]

Hence, the time ( in years ) = [tex]\frac{105.84}{12}=8.82[/tex]

Answer:

The future value of the annuity due to the nearest cent is $2956.

Step-by-step explanation:

Consider the provided information:

It is provided that monthly payment is $175, interest is 7% and time is 11 years.

The formula for the future value of the annuity due is:

[tex]FV of Annuity Due = (1+r)\times P[\frac{(1+r)^{n}-1}{r}][/tex]

Now, substitute P = 175, r = 0.07 and t = 11 in above formula.

[tex]FV of Annuity Due = (1+0.07)\times 175[\frac{(1+0.07)^{11}-1}{0.07}][/tex]

[tex]FV of Annuity Due = (1.07)\times 175[\frac{1.10485}{0.07}][/tex]

[tex]FV of Annuity Due = 187.5(15.7835)[/tex]

[tex]FV of Annuity Due = 2955.4789[/tex]

Hence, the future value of the annuity due to the nearest cent is $2956.

Answer:

The cutoff score should be 75.8 marks

Step-by-step explanation:

The cut-off should be set as the value corresponding to an area 15% in the normal distribution diagram.

Using the standard distribution tables we have value of standard normal deviate (Z) corresponding to area of 15% = -1.04

Thus we have

[tex]-1.04=\frac{X-\bar{X}}{\sigma }\\\\X=-1.04\times 5.5+81\\X=75.8[/tex]

Answer:

CF+PI=[tex]c_1e^{2x}+c_2e^{x}[/tex]+[tex]2e^{3t}[/tex]

Step-by-step explanation:

we have given y"-3y'=2y=[tex]4e^{3t}[/tex]

this differential equation solution have two part that CF and PI

CALCULATION OF CF :

[tex]m^2-3m+2=0[/tex]

[tex]m^2-2m-m+2=0[/tex]

[tex](m-1)(m-2)=0[/tex]

m=1 and m=2

so CF=[tex]c_1e^{2x}+c_2e^{x}[/tex]

CALCULATION OF PI :

PI =   [tex]\frac{4e^{3t}}{(m-1)(m-2)}[/tex]

at m= 3 in PI

[tex]PI=\frac{4e^{3t}}{2}=2e^{3t}[/tex]

so the complete solution is

CF+PI=[tex]c_1e^{2x}+c_2e^{x}[/tex]+[tex]2e^{3t}[/tex]

Answer with explanation:

Matrix A= (2 × 2) Matrix

Trace A= -9

Also,Determinant A= |A|=18

⇒Characteristics Polynomial is given by

Δ(A)=A² -A ×trace (A)+Determinant (A)

=A²+9 A+18

=(A+6)(A+3)

So, eigenvalues can be obtained by substituting :

 Δ(A)=0

(A+6)(A+3)=0

A= -6 ∧ A= -3

Two Eigenvalues are = -6, -3

Answer:

The answer is - budget constraint

Step-by-step explanation:

The slope of the budget constraint is determined by the relative price of the two goods, which is calculated by taking the price of one good and dividing it by the price of the other good.  

A budget constraint happens when a consumer demonstrates limited consumption patterns by a certain income.

The derivative of [tex]f(x,y,z)[/tex] at a point [tex]p_0=(x_0,y_0,z_0)[/tex] in the direction of a vector [tex]\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k[/tex] is

[tex]\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}[/tex]

We have

[tex]f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k[/tex]

and

[tex]\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}[/tex]

Then the derivative at [tex]p_0[/tex] in the direction of [tex]\vec a[/tex] is

[tex]3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}[/tex]

Answer:

Odd numbers that are multiple of 5 and are in between 22 and 66 are-

25, 35, 45, 55, 65

Let this set be represented by A

A= {25, 35, 45, 55, 65}

the above form represents the set in its roster form

The set notation for the odd numbers between 22 and 66 that are multiples of 5 is { x ∈ N | x is odd, 22 < x < 66, x ≡ 0 (mod 5) }.

The set notation for the odd numbers between 22 and 66 that are multiples of 5 can be written as:

{ x ∈ N | x is odd, 22 < x < 66, x ≡ 0 (mod 5) }

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The function log311y can be expressed as the sum of two logarithms, log3(11) + log3(y), according to the product rule of logarithms.

The function log311y represents the logarithm base 3 of the product of the numbers 11 and y. Using the rules of logarithms, we can rewrite this expression as a sum of two logarithms.

According to the product rule of logarithms, the logarithm of a product is the sum of the logarithms of the component numbers. Applying this rule to our expression, log311y becomes:

log3(11) + log3(y)

This is the sum of the logarithm base 3 of 11 and the logarithm base 3 of y. In conclusion, the function log311y can be expressed as the sum of the separate logarithms: log3(11) + log3(y).

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The logarithm log3(11y) can be broken down into two simpler logarithms, log3(11) and log3(y), by using properties of logarithms. This is the sum of the two simpler logarithms.

To express the logarithm log3(11y) as the sum or difference of logarithms, we will utilize the properties of logs:

Applying these properties to the given logarithm, we find:

log3(11y) = log3(11) + log3(y)

Thus, the original logarithm has been expressed as a sum of two simpler logarithms.

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The net force is what remains of the pull when we subtract the friction force:

[tex]F = 64-36 = 28N[/tex]

Now, use the law

[tex]F=ma[/tex]

and solve it for the acceleration

[tex]a = \dfrac{F}{m}[/tex]

to get the result:

[tex]a = \dfrac{28}{14}=2[/tex]

Answer:

2 m/s²

Step-by-step explanation:

F = applied force in the horizontal direction = 64 N

f = frictional force acting between the box and the floor = 36 N

m = mass of the box = 14 kg

a = acceleration of the box = ?

Force equation along the horizontal direction is given as

F - f = ma

Inserting the values

64 - 36 = 14 a

28 = 14 a

a =  [tex]\frac{28}{14}[/tex]

a = 2 m/s²

Answer:

y = -3/5x - 12/5

Step-by-step explanation:

The equation I'm going to give is going to be in slope-intercept form. If you need it in point-slope, I can do so in an edit or the comments.

Slope-intercept form is: y = mx + b where m is the slope, b is the y-intercept.

So let's plug in our given slope:

y = -3/5x + b

Using this, we now plug in our x- and y-coordinates from the given point to solve for b.

-3 = -3/5(1) + b

-3 = -3/5 + b

Add 3/5 to both sides to isolate variable b.

-3 + 3/5 = b

-15/5 + 3/5 = b

-12/5 = b

Plug this new info back into the original equation and your answer is

y = -3/5x - 12/5

Answer:

It's actually C

Step-by-step explanation:

don't forget about the probability of 0 too

you have to add the two probability formulas

The probability of observing, at most, one patient who will be cured of breast Cancer will be P = (20 / q²⁰) + 20 / (0.20 x 0.80¹⁹). Then the correct option is C.

Its basic premise is that something will almost certainly happen. The percentage of favorable events to the total number of occurrences.

A new drug on the market is known to cure 20% of patients with breast cancer.

p = 0.20

q = 1 – 0.20

q = 0.80

If a group of 20 patients is randomly selected.

The probability of observing, at most, one patient who will be cured of breast Cancer will be

P = (20 / q²⁰) + 20 / (0.20 x 0.80¹⁹)

Then the correct option is C.

More about the probability link is given below.

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The magnetic field H(z,t) of an electromagnetic wave is related to the electric field E(z,t) by a factor of the speed of light. Therefore, if E(z,t) = 4cos(6 x 10^8 t - 2z) +3 sin(6 x 10^8 t -2z), the associated magnetic field would be H(z,t) = (4/c) cos(6 x 10^8 t - 2z) +(3/c) sin(6 x 10^8 t -2z), where c is speed of light, approximately 3 x 10^8 m/s.

The question is asking for the associated magnetic field H(z,t) of an Electromagnetic wave given the electric field E(z,t). A crucial fact to know for this question is that the electric and magnetic fields in an electromagnetic wave are perpendicular to each other and the direction of propagation. They also have a constant ratio of magnitudes in free space or air, which is the speed of light given by c = 1/√εOMO. Because of these relations, we know that we can find the magnetic field by simply dividing the given electric field by the speed of light in units that match the given Electric field.

So, if E(z,t) = 4cos(6 x 10^8 t - 2z) +3 sin(6 x 10^8 t -2z), then the associated magnetic field would be H(z,t) = (4/c) cos(6 x 10^8 t - 2z) +(3/c) sin(6 x 10^8 t -2z), where c is the speed of light, approximately 3 x 10^8 m/s.

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To find the associated magnetic field H(z, t), you can use Faraday's law of electromagnetic induction. This law states that the rate of change of magnetic flux through a surface is equal to the induced electromotive force (EMF) along the boundary of the surface. By following a step-by-step process, you can find the magnetic field B(z, t) using the given electric field E(z, t).

The associated magnetic field H(z, t) can be found by using Faraday's law of electromagnetic induction. Faraday's law states that the rate of change of magnetic flux through a surface is equal to the electromotive force (EMF) along the boundary of the surface. In this case, the magnetic field is changing due to the time-dependent electric field, so we can use Faraday's law to find the magnetic field.

By following these steps, you can find the associated magnetic field H(z, t) using Faraday's law of electromagnetic induction.

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[tex]1.\rightarrow \frac{dy}{dx}-y=e^x y^2\\\\\rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=e^x\\\\ \text{put},\frac{-1}{y}=z\\\\ \frac{dy}{y^2} =d z\\\\ \frac{dy}{dx} \times \frac{1}{y^2}=\frac{dz}{dx}\\\\\frac{dz}{dx} +z=e^x\\\\ \text{Integrating factor}=e^{\int {1} \, dx}\\\\=e^x \\\\ \text{Multiplying both sides by }e^x\\\\e^x(\frac{dz}{dx} +z)=e^{2x}\\\\ \text{Integrating both sides}\\\\z e^x=\frac{e^{2x}}{2}+C\\\\ \frac{-e^x}{y}=\frac{e^{2x}}{2}+C[/tex]

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[tex]\rightarrow \frac{dy}{dx}=x+\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}-x=\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}+y(1-\frac{1}{x})=x\\\\\text{Integrating factor}=e^{\int{1-\frac{1}{x}}\,dx}\\\\=e^{x-\log x}\\\\ \text{Multiplying both sides by} e^{x-\log x}\\\\e^{x-\log x}\times[\frac{dy}{dx}+y(1-\frac{1}{x})]=x \times e^{x-\log x}\\\\y\times e^{x-\log x} =\int x \times e^{x-\log x} \, dx}\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{e^{\log x}}\,dx\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{x} \, dx\\\\y\times e^{x-\log x}=e^x+K[/tex]