pls pls pls pls help me !!!!! i’ll mark brainliest 50 points

(a) P{L1 = 0}: Since L1 is the last Poisson event time before m, and m is uniformly distributed on [0, 1], we have:

P{L1 = 0} = P{no Poisson events in [0, m]} = e^(-λm)

(b) P{L1 < x}, 0 < x < m: We have:

P{L1 < x} = P{there is at least one Poisson event in [0, x]} = 1 - P{no Poisson events in [0, x]} = 1 - e^(-λx)

(c) P{R1 = 1}: Since R1 is the first Poisson event time after m, we have:

P{R1 = 1} = P{no Poisson events in [m, 1]} = e^(-λ(1-m))

(d) P{R1 = x}, m < x < 1: We have:

P{R1 = x} = P{there is no Poisson event in [m, x]} * P{there is at least one Poisson event in [x, 1]}

= e^(-λ(x-m)) * (1 - e^(-λ(1-x)))

(e) E[R-L]: We have:

E[R-L] = E[(R1-L1) + (R2-L2) + ... + (Rn-Ln)]

= E[R1-L1] + E[R2-L2] + ... + E[Rn-Ln] (by linearity of expectation)

= nE[R1-L1] (since the n copies are identical)

To find E[R1-L1], note that R1-L1 represents the length of the fragment that contains the marked position. This length is distributed as an exponential random variable with parameter λ, since it is the time until the next Poisson event in a Poisson process with rate λ. Therefore:

E[R1-L1] = 1/λ

Thus, we have:

E[R-L] = n/λ

This means that as n (the number of copies) becomes large, the expected length of the interval containing the marked position becomes smaller and smaller, converging to 0 as n approaches infinity.

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Answer:5/6

Step-by-step explanation: Theres 6 people that you are splitting 5 oranges among. so you do the equation 5÷6 or 5/6 which is the answer your looking for

Answer:

5/6

Step-by-step explanation:

=0.8333.......................

The text is asking for a step-by-step sequence to solve quadratic problems using the quadratic formula.

A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one squared term. The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants and x is the variable. The quadratic formula is used to find the solution(s) of a quadratic equation. The formula is x = (-b ± sqrt(b² - 4ac)) / 2a.

Answer:

Step-by-step explanation:

put your equation into

ax²+bx+c

determine a, b, and c

plug into formula

simplify numbers under the square root first (b²-4ac)

then simplify the root.  ex.  √12 can be simplified to 2√3

then reduce if the bottom can be reduced with both of the terms on top

The height of the pillar is 14 feet.

We can use the ratio of the building's height to the length of its shadow to solve for the pillar's height and apply it to the pillar as well. This is because similar-shaped objects will produce shadows that are proportional to their size.

Let h represent the pillar's height, and let's establish a proportion:

height of building / length of building's shadow = height of pillar / length of pillar's shadow

Substituting the given values:

70 / 50 = h / 10

We can simplify this proportion by cross-multiplying:

70 x 10 = 50h

700 = 50h

And solving for h:

h = 700 / 50 = 14

Therefore, the height of the pillar is 14 feet.

In conclusion, we can solve for the pillar's height by establishing a ratio between the building's height and the length of its shadow and applying that ratio to the pillar.

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det(A) = ad - bc.

Matrix X is X = [8/5 -6/5, -2, 2/5 2/5]

We know that the adjugate of a 2x2 matrix is:

adj(A) = [d -b, -c a]

A = [a b, c d]

det(A) = ad - bc.

So, from the given adj(A), we have:

d - b = 1

-c = 2

-a = 1

d - c = 1

We have c = -2. Then, from the fourth equation, we have d = 1 - c = 3. Substituting these values into the first and third equations, we get:

b = 2

a = -1

So, the matrix A is:

A = [-1 2, -2 3]

We are given that AX = A + X. Substituting the matrix A and simplifying, we get:

AX = A + X

=> A(X - I) = X - A

=> (X - I)(-A) = X - A

=> X - I = (-A)⁻¹ (X - A)

=> X - I = A⁻¹ (X - A)

=> X = A⁻¹ X - A⁻¹ A + I

=> X = A⁻¹ (X - A) + I

Since we know A, we can find its inverse:

A⁻¹ = 1/(ad - bc) [d -b, -c a] = 1/5 [3 -2, 2 -1]

Substituting this into the above equation, we get:

X = 1/5 [3 -2, 2 -1] (X - [-1 2, -2 3]) + I

Simplifying this, we get:

X = 1/5 [4X + 5, -6X - 10, -2X - 10, 3X + 5]

Equating the corresponding elements on both sides, we get the following system of equations:

4x + 5 = x

-6x - 10 = 2

-2x - 10 = 1

3x + 5 = 3

Solving this system of equations, we get:

x = -1

Therefore, det(A) = ad - bc = (-1)(3) - (2)(-2) = -1 + 4 = 3.

And, matrix X is:

X = [8/5 -6/5, -2, 2/5 2/5]

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Therefore, the margin of error for a 95% confidence interval for the population mean of AAA batteries (with a population standard deviation of 0.5 hours and a population mean of 9 hours) based on a sample size of 25 batteries is 0.196 hours.

What is margin error?

Margin of error is a statistical term that refers to the amount of error that is allowed in a survey or poll's results. It is a measure of the precision of the results and indicates the range within which the true value is likely to fall.

In statistical terms, the margin of error is calculated as a percentage of the total number of respondents in a survey, and it takes into account factors such as the size of the sample, the level of confidence desired, and the variability in the responses.

For example, if a poll has a margin of error of +/- 3%, it means that the actual value of the result could be up to 3% higher or 3% lower than the reported value in the survey.

The formula for the margin of error for a 95% confidence interval is:

Margin of error = z * (σ / [tex]\sqrt(n)[/tex])

Where:

z = the z-score for the desired confidence level (95% confidence level corresponds to z = 1.96)

σ = the population standard deviation

n = the sample size

Substituting the given values into the formula:

Margin of error = 1.96 * (0.5 /[tex]\sqrt(25)[/tex])

Margin of error = 1.96 * (0.5 / 5)

Margin of error = 1.96 * 0.1

Margin of error = 0.196

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No, W is not a subspace of R3 because it is not closed under vector addition and scalar multiplication, even though it contains the zero vector.

A set must meet three requirements to be a subspace of a vector space: (1) it must include the zero vector, (2) it must be closed under vector addition, and (3) it must be closed under scalar multiplication.

While W includes the zero vector (0, 0, 0), vector addition does not close it. For example, the triples (3, 4, 5) and (5, 12, 13) are both Pythagorean, but their addition (8, 16, 18) is not. As a result, W does not meet the second requirement and is not a subspace of R3.

Under scalar multiplication, W is likewise not closed. When we multiply the Pythagorean triple (3, 4, 5) by -1, we obtain (-3, -4, -5), which is not a Pythagorean triple. Therefore, W does not satisfy the third condition and is not a subspace of R.

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The solution of inequality is -9 < x < 12.

The correct option is B.

We have,

-2 < x/3 + 1 < 5

Now, solving the inequation in parts as

-2 < x/3 + 1

-2 - 1 < x/3

-3 < x/3

-9 < x

and, x/3 + 1 < 5

x/3 < 4

x <12

Thus, the solution of inequality is -9 < x < 12.

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Answer: 87

Step-by-step explanation: 3 x 29=87 because you are multiplying the amount of people in the survey by 3

The singular, affirmative and formal command for each of the following are as follows: 1.tener: tenga, 2.conocer: conozca 3.buscar:busque 4.ir:vaya 5.ser:sea

What Spanish Affirmative and Negative commands?

The indicative, subjunctive, and imperative verb moods are the three primary categories of verb moods in Spanish.

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The statement "consider a wave form s(t)=5 sin 10 π t 2 sin 12 π t. the signal s(t) is sampled at 10 hz. do you expect to see aliasing" is true because aliasing is expected.

When sampling a signal s(t) = 5 sin(10πt) * 2 sin(12πt) at 10 Hz, you can expect to see aliasing. The Nyquist sampling theorem states that a signal should be sampled at least twice the highest frequency present in the signal to avoid aliasing.

The two sinusoids in s(t) have frequencies of 5 Hz (10πt) and 6 Hz (12πt). The highest frequency is 6 Hz, so according to the Nyquist theorem, the signal should be sampled at least at 12 Hz (2 times the highest frequency) to avoid aliasing. Since the signal is sampled at 10 Hz, which is lower than the required 12 Hz, aliasing will occur.

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If the curve x2/3+y2/3=4 and Let L be the tangent line to this curve at the point (1,3√3), and let A and B be the x- and y-intercepts of L, then the length of line segment AB is √38 (option b).

Explanation:

To find the length of line segment AB, follow these steps:

Step 1: We need to first find the equation of tangent line L at the given point (1, 3√3) on the curve x^(2/3) + y^(2/3) = 4.

Step 1. Find the derivative dy/dx using implicit differentiation:
(2/3)x^(-1/3) + (2/3)y^(-1/3)(dy/dx) = 0

Step 2. Solve for dy/dx (the slope of tangent line L) at point (1, 3√3):
(2/3)(1)^(-1/3) + (2/3)(3√3)^(-1/3)(dy/dx) = 0
dy/dx = -1/2

Step 3. Use the point-slope form to find the equation of tangent line L:
y - 3√3 = -1/2(x - 1)

Step 4. Find the x- and y-intercepts (A and B) of line L:
x-intercept (A): y = 0
0 - 3√3 = -1/2(x - 1)
x = 2 + 6√3

y-intercept (B): x = 0
y - 3√3 = -1/2(0 - 1)
y = 3√3 + 1/2

Step 5. Calculate the length of the line segment AB using the distance formula:
AB = √[(2 + 6√3 - 0)^2 + (3√3 + 1/2 - 0)^2] = √38

The length of line segment AB is √38 (option b).

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z = 2√13(cos 5.49 + i sin 5.49) in polar form.

We can use Euler's formula, e^(iθ) = cos(θ) + i sin(θ), to write complex numbers in polar form.

1. For e^(iπ/3), we have:

e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + i√3/2

Therefore, a = 1/2 and b = √3/2.

2. For e^(i4π/3), we have:

e^(i4π/3) = cos(4π/3) + i sin(4π/3) = -1/2 - i√3/2

Therefore, a = -1/2 and b = -√3/2.

3. For z = 6 - 4i, we can find the magnitude (or modulus) of z, r, using the Pythagorean theorem:

|r| = √(6^2 + (-4)^2) = √52 = 2√13

To find the angle, theta, we can use the inverse tangent function:

tan⁻¹(-4/6) = -tan⁻¹(2/3) ≈ -0.93 radians

However, since we want the angle to satisfy 0 ≤ θ < 2π, we need to add 2π to the angle if it is negative:

θ = 2π - 0.93 ≈ 5.49 radians

Therefore, z = 2√13(cos 5.49 + i sin 5.49) in polar form.

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The corresponding z-score for a random variable x that is normally distributed with a mean of 395 and a standard deviation of 23, when  x = 35 is:  -15.65

To find the z-score, you can use the following formula:

z = (x - μ) / σ

where z is the z-score, x is the value of the random variable, μ is the mean, and σ is the standard deviation.

Step 1: Identify the values.
x = 35, μ = 395, and σ = 23.

Step 2: Substitute the values into the formula.
z = (35 - 395) / 23

Step 3: Calculate the z-score.
z = (-360) / 23 = -15.65

So, the corresponding z-score for x = 35 is approximately -15.65.

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Answer:

Step-by-step explanation:

To find the average value of the function f(x) = x^2(x^3 + 30) on the interval [-3, 3], we use the formula:

fave = (1/(b-a)) * ∫[a,b] f(x) dx

where a = -3 and b = 3.

So, we have:

fave = (1/(3-(-3))) * ∫[-3,3] x^2(x^3 + 30) dx

fave = (1/6) * [∫[-3,3] x^5 dx + 30∫[-3,3] x^2 dx]

fave = (1/6) * [0 + 30(2*3^3)]

fave = 2430

Therefore, the average value of the function f on the given interval is 2430.

To find the average value of the function h(u) = In(u) on the interval [1, 5], we use the formula:

have = (1/(b-a)) * ∫[a,b] h(u) du

where a = 1 and b = 5.

So, we have:

have = (1/(5-1)) * ∫[1,5] ln(u) du

have = (1/4) * [u ln(u) - u] from 1 to 5

have = (1/4) * [(5 ln(5) - 5) - (ln(1) - 1)]

have = (1/4) * (5 ln(5) - 4)

have = 0.962

Therefore, the average value of the function h on the given interval is approximately 0.962.

To find all numbers b such that the average value of f(x) = 6 + 10x - 9x^2 on the interval [0, b] is equal to 7, we use the formula:

fave = (1/(b-a)) * ∫[a,b] f(x) dx

where a = 0 and b = b.

So, we have:

7 = (1/b) * ∫[0,b] (6 + 10x - 9x^2) dx

7b = [6x + 5x^2 - 3x^3/3] from 0 to b

7b = 2b^2 - 3b^3/3 + 6

21b = 6b^2 - b^3 + 18

b^3 - 6b^2 + 21b - 18 = 0

Using synthetic division, we find that b = 2 is a root of this polynomial equation. Dividing by (b-2), we get:

(b-2)(b^2 - 4b + 9) = 0

The quadratic factor has no real roots, so the only solution is b = 2.

Therefore, the only number b such that the average value of f(x) on the interval [0, b] is equal to 7 is 2.

To estimate the average world population to the nearest million during the time period of 1950 to 1980, we need to find:

ave = (1/(1980-1950)) * ∫[1950,1980] P(t) dt

ave = (1/30) * ∫[1950,1980] 2560e^(0.017185t) dt

Using the formula for integrating exponential functions, we get:

ave = (1/30) * [2560

Answer:

In this problem it is saying that it is $35 less than the original cost so to find this you need to add. This should be just $149 + $35 to solve it which is a total of $184 which is the original cost for the phone.

$184 is your answer

Step-by-step explanation:

$114

as the question says the price is $35 less than the original cost, which means 149-35 which equal 144.

Answer:

The statements that are true are:

6 + 3 = 9 and 4 · 4 = 16 (both are true statements)

5 · 3 = 15 or 7 + 5 = 20 (at least one of these statements is true, since the word "or" means that only one of the two statements needs to be true for the entire statement to be true)

The other two statements are false:

If 6 > 10, then 8 · 3 = 24 (this statement is false, because the premise "6 > 10" is false, and a false premise can never imply a true conclusion)

If 6 · 3 = 18, then 4 + 8 = 20 (this statement is false, because the conclusion "4 + 8 = 20" does not follow logically from the premise "6 · 3 = 18")

The slope of the parametric curve x=-4t^2-4, y=6t^3 at the point corresponding to t is -9t/4.

To find the slope of the parametric curve x=-4t^2-4, y=6t^3 at the point corresponding to t, follow these steps:
1.  Find the derivatives of both x and y with respect to t:
   dx/dt = -8t
   dy/dt = 18t^2
2. The slope of the parametric curve is the ratio of the derivatives, dy/dx.

    To find this, divide dy/dt by dx/dt:

    dy/dx = (dy/dt) / (dx/dt)

              = (18t^2) / (-8t)
3. Simplify the expression:
   dy/dx = -9t / 4
So, the slope of the parametric curve x=-4t^2-4, y=6t^3 at the point corresponding to t is -9t/4.

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Answer:

-1(-4)=-5

Step-by-step explanation:

as h = -1

let's firstly get the EQUATion, of the graph before we get the inequality.

so we have a quadratic with two zeros, at -6 and 8, hmmm and we also know that it passes through (-2 , 10)

[tex]\begin{cases} x = -6 &\implies x +6=0\\ x = 8 &\implies x -8=0\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{original~polynomial}{a ( x +6 )( x -8 ) = \stackrel{0}{y}}\hspace{5em}\textit{we also know that } \begin{cases} x=-2\\ y=10 \end{cases}[/tex]

[tex]a ( -2 +6 )( -2 -8 ) = 10\implies a(4)(-10)=10\implies -40a=10 \\\\\\ a=\cfrac{10}{-40}\implies a=-\cfrac{1}{4} \\\\[-0.35em] ~\dotfill\\\\ -\cfrac{1}{4}(x+6)(x-8)=y\implies -\cfrac{1}{4}(x^2-2x-48)=y \\\\\\ ~\hfill {\Large \begin{array}{llll} -\cfrac{x^2}{4}+\cfrac{x}{2}+12=y \end{array}}~\hfill[/tex]

now, hmmm let's notice something, the line of the graph is a solid line, that means the borderline is included in the inequality, so we'll have either ⩾ or ⩽.

so hmmm we could do a true/false region check by choosing a point and shade accordingly, or we can just settle with that, since the bottom is shaded, we're looking at "less than or equal" type, or namely ⩽, so that's our inequality

[tex]{\Large \begin{array}{llll} -\cfrac{x^2}{4}+\cfrac{x}{2}+12\geqslant y \end{array}}[/tex]

Answer:

B) 11

Step-by-step explanation:

PEMDAS=parenthesis, exponents, multiplication, division, addition, and subtraction.

First start with the parenthesis:

20 x 1/4=5

9 + (5) - 6 ÷ 2

We don't have exponents or multiplication, so go onto division:

-6 ÷ 2 = -3

9+5-3

Finally addition and subtraction:

9+5-3=11

Hope this helps!

In a Nyquist plot, If G(s) has one pole in the right-half s plane, the system is marginally stable.

If G(s) has no pole in the right-half s plane, but has one zero in the right-half s plane, the system is stable.

In a Nyquist plot, the stability of a system can be determined by examining the number of encirclements of the -1 point. If the number of encirclements is equal to the number of right-half plane poles, then the system is unstable. If the number of encirclements is less than the number of right-half plane poles, then the system is marginally stable. If the number of encirclements is greater than the number of right-half plane poles, then the system is stable.

In the first case where G(s) has one pole in the right-half s plane, the Nyquist plot will encircle the -1 point once in the clockwise direction. Therefore, the number of encirclements is less than the number of right-half plane poles, which means the system is marginally stable.

In the second case where G(s) has one zero in the right-half s plane, the Nyquist plot will not encircle the -1 point at all. Therefore, the number of encirclements is less than the number of right-half plane poles, which means the system is stable.

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Based on the analysis, there is no 3x3 matrix A such that A^2 = -I_3. To understand this analysis let's consider whether there exists a 3x3 matrix A such that A^2 = -I_3, where I_3 is the 3x3 identity matrix.

Step:1. Start by assuming that there is a 3x3 matrix A such that A^2 = -I_3.
Step:2. Recall that the determinant of a matrix squared (det(A^2)) is equal to the determinant of the matrix (det(A)) squared: det(A^2) = det(A)^2.
Step:3. Compute the determinant of both sides of the equation A^2 = -I_3: det(A^2) = det(-I_3).
Step:4. For the 3x3 identity matrix I_3, its determinant is 1. Therefore, the determinant of -I_3 is (-1)^3 = -1.
Step:5. From step 2, we know that det(A^2) = det(A)^2. Since det(A^2) = det(-I_3) = -1, we have det(A)^2 = -1.
Step:6. However, no real number squared can equal -1, which means det(A)^2 cannot equal -1.
Based on the analysis, there is no 3x3 matrix A such that A^2 = -I_3.

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The shape and size of the figure are unchanged. and other soltuions are below

The rule for translating a point 4 units left and 8 units up is (x, y) → (x - 4, y + 8).

Using the above rule

After the translation, the new coordinates of A are (3, 13).

The rule for reflecting a point over the y-axis is (x, y) → (-x, y).

Using the above rule

After the reflection, the coordinates of A become (-7, 5).

The final figure is congruent to the original figure because translation and reflection are both rigid transformations, which preserve distance and angles between points.

Therefore, the shape and size of the figure are unchanged.

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The watch gains 4 min till 8:00 PM in the next evening and show 8:04 pm the next evening.

Considering that,

A broken watch adds ten seconds per hour.

Find the number of hours between 8:00 PM this evening and 8:00 PM the following evening.

There are 24 hours in a day.

number of seconds the defective watch gained.

1 hour equals 10 seconds

24 hours ÷ by 10

24 * 10 is 240 seconds.

Now figure out how many minutes your defective watch has gained.

60 s = 1 minute

240 sec = 240/60

= 4 min

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Missing parts;

A faulty watch gains 10 seconds an hour. If it is set correctly at 8:00 pm one evening, what time will it show when the correct time is 8:00 pm the following evening

Answer:

this is a correct answer 5√6/2

The resultant matrix represents the vertices of the translated triangle, with A' at (-3, 7), B' at (-2, 0), and C' at (1, 7).

A transformation matrix is a mathematical matrix that represents a geometric transformation of space.

To translate the triangle by (-2, 4), we need to add -2 to the x-coordinates and add 4 to the y-coordinates of each vertex. The transformation matrix for this translation is:

[tex]\left[\begin{array}{ccc}1&0&-2\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0&1&4\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0&0&1\end{array}\right][/tex]

To apply this transformation to the vertices of the triangle, we can represent each vertex as a column vector with a third coordinate of 1:

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

[tex]B=\left[\begin{array}{ccc}0&-4&1\end{array}\right][/tex]

[tex]C=\left[\begin{array}{ccc}3&3&1\end{array}\right][/tex]

To apply the translation, we multiply each vertex by the transformation matrix:

[tex]A'=\left[\begin{array}{ccc}1&0&-2\end{array}\right] \left[\begin{array}{ccc}-1&3&1\end{array}\right] \left[\begin{array}{ccc}-3&7&1\end{array}\right][/tex]

      [0 1  4] × [ 0 -4  1] = [-2  0  1]

      [0 0  1]   [ 1  3  1]   [-1  7  1]

[tex]B'=\left[\begin{array}{ccc}1&0&-2\end{array}\right] \left[\begin{array}{ccc}0&-4&1\end{array}\right] \left[\begin{array}{ccc}-2&-4&1\end{array}\right][/tex]

      [0 1  4] × [ 0 -4  1] = [-2   0  1]

      [0 0  1]   [ 0  3  1]   [-2   7  1]

[tex]C'=\left[\begin{array}{ccc}1&0&-2\end{array}\right] \left[\begin{array}{ccc}3&3&1\end{array}\right] \left[\begin{array}{ccc}1&7&1\end{array}\right][/tex]

      [0 1  4] × [ 3  3  1] = [ 1  7  1]

      [0 0  1]   [ 3  3  1]   [ 1  7  1]

The resultant matrix represents the vertices of the translated triangle, with A' at (-3, 7), B' at (-2, 0), and C' at (1, 7).

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The final image is real since it is located to the right of the second lens. It is also upright since the magnification is positive.

(a) Using the thin lens formula, 1/f = 1/p + 1/q, where f is the focal length, p is the object distance, and q is the image distance, we have:

1/f = 1/p - 1/q1

Substituting the given values, we get:

1/11.6 = 1/16.2 - 1/q1

Solving for q1, we get:

q1 = 6.97 cm

Therefore, the image formed by the first lens is located 6.97 cm to the right of the lens.

(b) The image formed by the first lens acts as an object for the second lens. Using the thin lens formula again, we have:

1/f = 1/p2 + 1/q1

Substituting the given values, we get:

1/5 = 1/p2 + 1/6.97

Solving for p2, we get:

p2 = 3.32 cm

The distance from the second lens to the image of the first lens is then:

q2 = p2 + q1 = 3.32 + 6.97 = 10.29 cm

Therefore, the image of the first lens is located 10.29 cm to the right of the second lens.

(c) The object distance for the second lens is simply the image distance of the first lens:

p2 = q1 = 6.97 cm

(d) Using the thin lens formula again, we have:

1/f = 1/p2 + 1/q2

Substituting the given values, we get:

1/5 = 1/6.97 + 1/q2

Solving for q2, we get:

q2 = 13.95 cm

Therefore, the final image is located 13.95 cm to the right of the second lens.

(e) The magnification of the first lens is given by:

m1 = -q1/p = -6.97/16.2 ≈ -0.43

(f) The magnification of the second lens is given by:

m2 = -q2/p2 = -13.95/3.32 ≈ -4.20

(g) The total magnification of the system is given by the product of the magnifications of the two lenses:

m = m1 × m2 ≈ 1.81

(h) The final image is real since it is located to the right of the second lens. It is also upright since the magnification is positive.

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