Suppose X ~ Exp(lambda) and Y = ln X. Find the probability density function of Y.

Approximately 25% of the tomatoes weigh less than, given the standard deviation and mean, A. 93 g.

25 % of tomatoes is the value that we are looking for. On the z - distribution table, the closest to this amount is 0.2486, and this has a z - score of - 0. 674.

With this z - score, we can use the z - score formula to find the amount that the 25 % of tomatoes weigh:

z = (x - μ) / σ

-0. 674 = (x - 95 ) / 3.6

x - 95 = -0. 674 x 3. 6

x = 95 - 2. 4264

x = 92. 5736

x = 93 grams

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The total amount of interest that Minh will need to pay the bank at the end of 5 years is $1,350 whose principal amount is $3,000

The formula for simple interest is:

I = P × r ×  t

Where:

I is the amount of interest

P is the principal amount borrowed

r is the interest rate per year

t is the time period in years

In this case, P = $3,000, r = 9% = 0.09 (as a decimal), and t = 5 years. Substituting these values into the formula, we get:

I = 3000 × 0.09 × 5

I = $1,350

Therefore, the total amount of interest that Minh will need to pay the bank at the end of 5 years is $1,350.

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The total amount of interest that Minh will need to pay the bank at the end of 5 years is $1,350 whose principal amount is $3,000

The formula for simple interest is:

I = P × r ×  t

Where:

I is the amount of interest

P is the principal amount borrowed

r is the interest rate per year

t is the time period in years

In this case, P = $3,000, r = 9% = 0.09 (as a decimal), and t = 5 years. Substituting these values into the formula, we get:

I = 3000 × 0.09 × 5

I = $1,350

Therefore, the total amount of interest that Minh will need to pay the bank at the end of 5 years is $1,350.

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To sketch the straight-line Bode plot of the gain only for the voltage transfer function T(S) = 20s/ S2 + 58s + 400, we first need to break it down into its constituent parts. The numerator is simply a constant gain of 20, while the denominator can be factored into two second-order terms:

T(S) = 20s/ (S+20)(S+20)

Using the standard Bode plot rules for second-order systems, we can plot each term separately and then combine them to get the overall plot. For each term, we need to find the resonant frequency, damping ratio, and gain at low and high frequencies.

For the first term (S+20), the resonant frequency is 20, the damping ratio is 1/2, and the low-frequency gain is 0 dB. At high frequencies, the gain rolls off at a rate of -20 dB/decade.

For the second term (S+20), the resonant frequency is also 20, the damping ratio is 1/2, and the low-frequency gain is 0 dB. However, at high frequencies, the gain rolls off at a rate of -40 dB/decade due to the double pole.

To combine these two plots, we simply add the gains at each frequency and use the steeper roll-off rate for the second term. The result is a straight-line Bode plot with a gain of 20 dB at low frequencies, a resonant peak at 20 rad/s, and a steep roll-off at high frequencies.

The plot will cross the 0 dB line at two points, one before and one after the resonant peak, due to the double pole in the transfer function.

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The domain of e(h) is given as C. the real numbers from 0 to 24

The domain of e(h) should be the set of values that h can take on. Since h represents the number of hours of light per day that the chickens are exposed.,

Therefore, the domain would be the real numbers from 0 to 24. So, the correct answer is: C. the real numbers from 0 to 24

With this in mind, it can be seen that the correct answer is option C

Real numbers encompass all logical and irrational figures, visible on a numerical graph with three values: positive, negative, or zero.

These numbers offer pertinent traits as they allow their arithmetic to undergo operations like adding, subtracting, multiplying, and dividing without loss of significance.

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Answer:

224

Step-by-step explanation:

Total marks = x

x * 25% + 64 = x * 40% - 32

x * 15% = 96

x = 640

Maximum Marks = 640..

Marks Scored = 25% of 640

                        = 160
Marks Required to pass = 160 + 64

                                        = 224

The International Society for Automation (ISA) promotes that we must treat fairly and respectfully all colleagues and co-workers, while also "recognizing each of their unique contributions and individual capabilities full of strengths and weaknesses".Therefore option A is correct.

The International Society for Automation (ISA) promotes that we must treat fairly and respectfully all colleagues and co-workers, while also recognizing each of their unique contributions and individual capabilities full of strengths and weaknesses.

ISA emphasizes constructive and supportive feedback to help aid in the development of employees, rather than judging their quality and/or quantity of work critically and rashly.

Providing financial support or other incentives may be helpful in boosting motivation, but ultimately, treating colleagues fairly and respectfully is key to fostering a positive work environment.

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The International Society for Automation (ISA) promotes that we must treat fairly and respectfully all colleagues and co-workers, while also "recognizing each of their unique contributions and individual capabilities full of strengths and weaknesses".Therefore option A is correct.

The International Society for Automation (ISA) promotes that we must treat fairly and respectfully all colleagues and co-workers, while also recognizing each of their unique contributions and individual capabilities full of strengths and weaknesses.

ISA emphasizes constructive and supportive feedback to help aid in the development of employees, rather than judging their quality and/or quantity of work critically and rashly.

Providing financial support or other incentives may be helpful in boosting motivation, but ultimately, treating colleagues fairly and respectfully is key to fostering a positive work environment.

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Note that if x1 = x2 = x3 = 0, then the constraint is satisfied regardless of the values of x4 and x5.

a. The constraint for choosing no fewer than three projects can be written as:

x1 + x2 + x3 + x4 + x5 ≥ 3

b. The constraint for selecting project 4, if project 3 is chosen, can be written as:

x3 ≤ x4

Note that if x3 = 0, then the constraint is satisfied regardless of the value of x4.

c. The constraint for not selecting Project 5 if Project 1 is chosen can be written as:

x1 + x5 ≤ 1

Note that if x1 = 0, then the constraint is satisfied regardless of the value of x5.

d. The constraint for staying within the budget can be written as:

100x1 + 200x2 + 150x3 + 75x4 + 300x5 ≤ 450

e. The constraint for selecting no more than two of projects 1, 2, and 3 can be written as:

x1 + x2 + x3 ≤ 2

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The surface area of the larger pyramid is 16 times the surface area of the smaller pyramid.

To compare the surface areas of the two square pyramids, we first need to find the surface area of each pyramid. The surface area of a square pyramid is given by the formula SA = [tex]L^2[/tex] + 2L√([tex]L^2[/tex]/4 + [tex]H^2[/tex]).

Where,

L is the length of one side of the base and

H is the height of the pyramid.
For the larger pyramid, L = 10 inches and H = 5 inches. Plugging these values into the formula, we get:
SA = [tex]10^2[/tex] + 2(10)√([tex]10^2[/tex]/4 + [tex]5^2[/tex])
SA = 100 + 100√26
SA ≈ 272.94 square inches
For the smaller pyramid,

L = 2.5 inches and

H = 5 inches.

Plugging these values into the formula, we get:
SA = [tex]2.5^2[/tex] + 2(2.5)√([tex]2.5^2[/tex]/4 + [tex]5^2[/tex])
SA = 6.25 + 12.5√6
SA ≈ 41.23 square inches
Now we can compare the surface areas by dividing the surface area of the larger pyramid by the surface area of the smaller pyramid:
SA(larger) / SA(smaller) = 272.94 / 41.23
SA(larger) / SA(smaller) ≈ 6.62
Therefore, the surface area of the larger pyramid is approximately 6.62 times the surface area of the smaller pyramid.

In other words, the larger pyramid has a much larger surface area than the smaller pyramid.

This makes sense because the larger pyramid has a much larger base and height, which results in a much larger overall volume and surface area.

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The answer is (D) The limit cannot be determined from the information given.

To start, we need to simplify the given equation:

f(2) = g(2) + h(2)

Substituting 2 into g(x), we get:

g(2) = sin(2π/3 + 2) + 3

Using the unit circle, we can see that sin(2π/3 + 2) = sin(2π/3 - 1) = √3/2 * cos(1) - 1/2 * sin(1)

So, g(2) = √3/2 * cos(1) - 1/2 * sin(1) + 3

Now, substituting 2 into h(z), we get:

h(2) = -2/(2+3)

Simplifying, we get h(2) = -2/5

Therefore, f(2) = g(2) + h(2) = √3/2 * cos(1) - 1/2 * sin(1) + 3 - 2/5

Now, to find the limit as x approaches 3, we need to evaluate:

lim (x→3) f(x)

However, since we only have information for f(2), we cannot determine the limit as x approaches 3.

Therefore, the answer is (D) The limit cannot be determined from the information given.

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In conclusion, we have shown that every element of Sn can be written as a product of transpositions of the form (1, k) for 2 ≤ k ≤ n.

To show that every element of Sn can be written as a product of transpositions of the form (1, k) for 2 ≤ k ≤ n, we need to prove that any permutation in Sn can be expressed as a product of such transpositions.

Firstly, note that it is enough to show that this is true for transpositions because any permutation in Sn can be expressed as a product of transpositions.

Now, consider a permutation π in Sn. We can write π as a product of transpositions as follows:

π = (1, π(1))(1, π(2))...(1, π(n-1))

To see why this works, consider the effect of the first transposition (1, π(1)) on π. This transposition swaps 1 and the element π(1) in π. Then, consider the effect of the second transposition (1, π(2)) on the result of the first transposition. This transposition swaps 1 and the element π(2), which may or may not be equal to π(1). Continuing this process, we eventually end up with the permutation π.

Note that each transposition (1, k) can be expressed as the product of three transpositions:

(1, k) = (1, 2)(2, k)(1, 2)

Therefore, any permutation in Sn can be expressed as a product of transpositions of the form (1, k) for 2 ≤ k ≤ n.

In conclusion, we have shown that every element of Sn can be written as a product of transpositions of the form (1, k) for 2 ≤ k ≤ n.

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Thus, the 4th coordinate of the rectangle with the given coordinates is found as : D(8, -5). Area of the rectangle ABCD = 21 sq. units.

The Pythagorean theorem serves as the foundation for the distance formula. A line connecting two sites of interest is the hypotenuse of a right triangle, and this particular line connects the two points of interest.

The neighbouring side is obtained by joining the x-coordinates of a two points in a horizontal line, whereas the opposing side is obtained by joining the y-coordinates.

d=√((x2 - x1)²+(y2 - y1)²)

Given data:

vertices of a rectangle ABCD -

A(5, 2), B(8, 2), C(5, -5)

Let the 4 the vertex be D(x,y).

Plot the coordinates on the graph.

Now, we know that - opposite sides of the rectangle are equal.

Thus,

AB = CD

From graph,D(x,y).

x - (5 + 3) = 8

y - (2 - 7) = -5

Thus, the 4th coordinate of the rectangle with the given coordinates is found as : D(8, -5).

Area of the rectangle ABCD = length x breadth

length = 2 + 5 = 7 units

width = 8 - 5 = 3 units

Area of the rectangle ABCD = 7 x 3

Area of the rectangle ABCD = 21 sq. units.

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The number of different fruit salads that can be made using up to three  different fruits is 41

The number of ways of selecting r objects from n unlike objects is given by combinations

ⁿCₓ = n! / ( ( n - x )! x! )

where

n = total number of objects

x = number of choosing objects from the set

Given data ,

For each possible number of fruits (1, 2, or 3), we can count the number of ways to choose that many fruits from the available options and then sum up the results

To make a fruit salad with one fruit, we can choose any of the six available fruits. There are 6 ways to do this

To make a fruit salad with two fruits, we can choose any two fruits from the six available options. This can be done using the combination formula

C(6,2) = 6! / (2! x (6-2)!) = 15

So, there are 15 ways to make a fruit salad with two fruits

And , To make a fruit salad with three fruits, we can choose any three fruits from the six available options. This can be done using the combination formula:

C(6,3) = 6! / (3! x (6-3)!) = 20

So , the total number of ways A = 6 + 15 + 20

A = 41 ways

Hence , the different fruit salads is A = 41

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Answer:

The statement "If a and b are integers and a > b, then lal > Ibl" is true.

Explanation:

The estimated coefficient for the total loans and leases to total assets ratio indicates that holding the total expenses to assets ratio constant, a one unit increase in the total loans and leases to assets ratio is associated with an increase in the odds of being financially weak by a factor of -14.18755183 + 79.963941181 + 9.1732146 = 74.949604981.

This means that as the ratio of total loans and leases to total assets increases, the odds of being financially weak also increase by a significant factor. It is important to note that other factors may also contribute to financial weakness and that the coefficient should be interpreted in conjunction with other relevant data and factors.

The estimated coefficient for the total loans and leases to total assets ratio in terms of the odds of being financially weak can be interpreted as follows:

Holding the total expenses/assets ratio constant, a one unit increase in the total loans and leases-to-assets ratio is associated with an increase in the odds of being financially weak by a factor of 9.1732146.

This means that for each unit increase in the loans and leases-to-assets ratio, the likelihood of the institution being financially weak increases by a factor of 9.1732146, assuming all other factors remain the same.

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The differentiation of dy/dx is (A) -3.215.

A function's derivative with respect to an independent variable can be used to define differentiation. Calculus differentiates to measure the function per unit change in the independent variable.

To solve the differential equation dy/dx=x*cos(x²), we need to integrate both sides with respect to x.

Integrating the right-hand side involves the substitution u = x², du/dx = 2x, so that cos(x²) dx = (1/2)cos(u) du.

Substituting u = x² and cos(u) = cos(x²) in the integral we have:

dy/dx = x*cos(x²)

=> dy = x*cos(x²)dx

=> ∫ dy = ∫ x*cos(x²)dx

=> y = (1/2) sin(x²) + C

where C is a constant of integration.

To determine the value of C, we use the initial condition y=-3 when x=0:

y = (1/2)sin(x²) + C

=> -3 = (1/2)sin(0) + C

=> C = -3

Using a calculator or a table of values for sine, we have sin(π²) ≈ 0.0247, so:

y = (1/2)sin(π²) - 3

=> y ≈ -2.9876

Therefore, the answer is closest to option A, -3.215.

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A particular solution is y_p(t) = -1/2sin(t) + 1/2cos(t), and the general solution is: y(t) = y_h(t) + y_p(t) = c1e^(t/2)cos(√3t/2) + c2e^(t/2)sin(√3t/2) - 1/2sin(t) + 1/2cos(t).

a. For the equation y'' - y' + y = sin(t), the characteristic equation is r^2 - r + 1 = 0, which has complex roots r = (1 ± i√3)/2. Therefore, the homogeneous solution is y_h(t) = c1e^(t/2)cos(√3t/2) + c2e^(t/2)sin(√3t/2).

To find a particular solution, we assume it has the form y_p(t) = Asin(t) + Bcos(t), where A and B are unknown constants. Taking the derivatives, we get y_p'(t) = Acos(t) - Bsin(t) and y_p''(t) = -Asin(t) - Bcos(t). Substituting these into the original equation, we get:

(-Asin(t) - Bcos(t)) - (Acos(t) - Bsin(t)) + Asin(t) + Bcos(t) = sin(t)

Simplifying, we get:

2B = 1

Therefore, B = 1/2. Substituting this into the equation above, we get:

-Acos(t) + 1/2sin(t) + Acos(t) - 1/2sin(t) = sin(t)

Simplifying, we get:

A = -1/2

Therefore, a particular solution is y_p(t) = -1/2sin(t) + 1/2cos(t), and the general solution is:

y(t) = y_h(t) + y_p(t) = c1e^(t/2)cos(√3t/2) + c2e^(t/2)sin(√3t/2) - 1/2sin(t) + 1/2cos(t).

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The differential equation for the balance B in an investment fund with time, t, measured in years A) dB/dt= 0.05B - 11,000.

To write a differential equation for the balance B in an investment fund with time, t, measured in years, we can use the following information:

- The balance is earning interest at a continuous rate of 5% per year
- Payments are being made out of the fund at a continuous rate of $11,000 per year

Let's start by considering the interest earned on the balance. At a continuous rate of 5% per year, the interest earned can be expressed as 0.05B (where B is the balance). This represents the rate of change of the balance due to interest.

Now let's consider the payments being made out of the fund. At a continuous rate of $11,000 per year, the payments can be expressed as a constant rate of change of -11,000.

Putting these two rates of change together, we can write the differential equation for the balance B as:

dB/dt = 0.05B - 11,000

This equation represents the balance B as a function of time t, with the rate of change of B being equal to the interest earned (0.05B) minus the payments being made (-11,000) at any given time t.

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The differential equation for the balance B in an investment fund with time, t, measured in years A) dB/dt= 0.05B - 11,000.

To write a differential equation for the balance B in an investment fund with time, t, measured in years, we can use the following information:

- The balance is earning interest at a continuous rate of 5% per year
- Payments are being made out of the fund at a continuous rate of $11,000 per year

Let's start by considering the interest earned on the balance. At a continuous rate of 5% per year, the interest earned can be expressed as 0.05B (where B is the balance). This represents the rate of change of the balance due to interest.

Now let's consider the payments being made out of the fund. At a continuous rate of $11,000 per year, the payments can be expressed as a constant rate of change of -11,000.

Putting these two rates of change together, we can write the differential equation for the balance B as:

dB/dt = 0.05B - 11,000

This equation represents the balance B as a function of time t, with the rate of change of B being equal to the interest earned (0.05B) minus the payments being made (-11,000) at any given time t.

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The lengths of the legs of the right triangle are given as follows:

2.39 feet and 4.39 feet.

The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.

The theorem is expressed as follows:

c² = a² + b².

In which:

c is the length of the hypotenuse.

a and b are the lengths of the other two sides (the legs) of the right-angled triangle.

The parameters for this problem are given as follows:

Hence the value of x is obtained as follows:

x² + (x + 2)² = 5²

x² + x² + 4x + 4 = 25

2x² + 4x - 21 = 0.

Using a calculator, the positive solution for x is given as follows:

2.39.

Hence the legs are:

2.39 feet and 4.39 feet.

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The volume of the rectangular prism shown is 12 3/4 cubic feet. The correct option is (C).

The volume of a rectangular prism is given by:

V = length x width x height

Given

height (h) = 4 1/4 ft = 17/4

width (w) 1 1/2 ft = 3/2

length (l)  = 2 ft.

Substitute the values:

Volume = length x width x height

= 2 ft x 3/2 ft x 17/4 ft

= 51/4 cubic feet

= 12 3/4 cubic feet

Therefore, the volume of the rectangular prism is 12 3/4 cubic feet.

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a. We have shown that α + 1 = β.

b. We have shown that α² = β.

A substance that cannot be broken down into another substance is referred to as an element. Because each element is made up of its own type of atom, each element is distinct and distinct from the others.

Since K is a field with 4 elements, the nonzero elements of K form a cyclic group of order 3 under multiplication. Let us call this group G. Since G is cyclic, it has a unique generator, say g. Then we can write G = {1, g, g²}.

Now, since α and β are both in K but not in Z2, they must be elements of G. Moreover, since K contains Z2 as a sub-field, α and β must be roots of the polynomial x² + x + 1 over Z2. Therefore, we have the following possibilities for α and β:

α = g

β = g²

or

α = g²

β = g

a. Let us first consider the case where α = g and β = g². Then we have:

α + 1 = g + 1

= g² + g + 1 (since g³ = 1)

= β + α + 1 (since β = g² and α = g)

= β

Therefore, we have shown that α + 1 = β.

b. Next, we will show that α² = β. Using the same assumption that α = g and β = g², we have:

α² = g²

= β

Therefore, we have shown that α² = β.

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A) y = x + 3 represents a straight line with a slope of 1 passing through the point (0,3).

B) y = -x + 3 represents a straight line with a slope of -1 passing through the point (0,3).

C) y = |x| + 3 represents two lines; one with a slope of 1 passing through the point (-3,0) and another with a slope of -1 passing through the point (3,0).

The points (x,y) that satisfy all three equations must lie on the lines with slope 1 and -1 passing through the point (0,3).

These lines intersect at the point (1,4) and the set of points that satisfy all three equations is the single point (1,4).

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Answer:

To convert gallons to liters, we need to multiply the number of gallons by 3.78541, which is the conversion factor.

Capacity in liters = 16 gallons * 3.78541 liters/gallon

Capacity in liters = 60.56 liters (rounded to the nearest tenth)

Therefore, the capacity of the gas tank in liters is approximately 60.6 liters.

Answer:

60.6

Step-by-step explanation:

First, we find how many liters there are in a gallon. We find there are 3.78541178 liters in a gallon. 16*3.78541178 =60.5665885 Rounding to the nearest tenth, we get 60.6 as our answer.

Sarah's profit-maximizing amount of output is  200 Sandwiches per day.

At the intersection of marginal cost (MC) and the demand curve, a firm will be producing the level of output where it maximizes its profit. This point is also known as the profit-maximizing point or the point of allocative efficiency.

For the given diagram or graph, Sarah's profit-maximizing amount of output will occur at the intersection of the marginal cost curve and the demand curve.

This point = $8 and 200 Sandwiches per day.

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Answer:

$546

Step-by-step explanation:

It is given that the variables:

a = adults

c = students.

There are 6 adults, and 38 students. Plug in 6 for a, and 38 for c in the given expression:

[tex]15a + 12c\\15(6) + 12(38)[/tex]

Simplify. Remember to follow PEMDAS. PEMDAS is the order of operations, and stands for:

Parenthesis

Exponents (& Roots)

Multiplications

Divisions

Additions

Subtractions

~

First, multiply 15 & 6, and 12 & 38 together:

[tex](15 * 6) + (12 * 38)\\(90) + (456)[/tex]

Next, simplify by adding:

[tex]90 + 456 = 546[/tex]

$546 is your total cost.

~

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The area of the irregular shape that is given above would be = 1135.89m²

To calculate the area of the given shape, the irregular shapes first divided into two and the separate areas calculated and added together.

For shape 1 ( triangle)

The formula for the area of a triangle = 1/2base ×height

where;

base = 9.3+23.7+9.3 = 42.3

height = 49.8-32.4 = 17.4

area = 1/2×42.3×17.4

= 736.02/2 = 368.01

Shape 2

The formula for the area of rectangle;

= Length×width

= 23.7×32.4

=767.88

Therefore the area of the irregular shape

= 368.01+ 767.88

= 1135.89m²

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The answer is non-conventional cash flow .

Cash flow refers to the movement of money in and out of a business or individual's financial accounts over a specific period of time. It represents the inflow and outflow of actual cash, as opposed to accounting profit or loss, which may include non-cash items such as depreciation or accruals .

When the algebraic signs on the net cash flows change more than once, the cash flow sequence is called non-conventional cash flow .

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The equation 78x45 + 22x45 can be simplified using the distributive property of multiplication over addition as follows:

The distributive property of multiplication over addition is a mathematical property that allows us to break down a multiplication problem involving the sum of two or more numbers.

The distributive property states that multiplying a number by a sum of two or more numbers is the same as doing each multiplication separately, then adding the products.

78x45 + 22x45

= (78 + 22) x 45

= 100 x 45

= 4500

Therefore, 78x45 + 22x45 = 4500.

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The equation 78x45 + 22x45 can be simplified using the distributive property of multiplication over addition as follows:

The distributive property of multiplication over addition is a mathematical property that allows us to break down a multiplication problem involving the sum of two or more numbers.

The distributive property states that multiplying a number by a sum of two or more numbers is the same as doing each multiplication separately, then adding the products.

78x45 + 22x45

= (78 + 22) x 45

= 100 x 45

= 4500

Therefore, 78x45 + 22x45 = 4500.

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Answer:

2

Step-by-step explanation:

Rate = Total number of customers / Total time

Rate = 14/7

Rate = 2

Therefore, the customers were entering the store at a rate of 2 customers per minute.

Answer:

The rate of change: -1

Step-by-step explanation:

The rate of change is the slope.

The equation is in slope-intercept form y = mx + b

m = the slope

The equation y = -x + 5

We see the slope is -1.

So, the rate of change is -1.

For a marketing firm which is considering making up to three new hires.

a) The probability that the firm will make at least one hire is equals to the 0.90.

b) The expected value and the standard deviation of the number of hires are 1.57 and 0.78 respectively.

We have a marketing firm is considering making up to three new hires. Let's consider the a random variable X that represents the hiring of candidates. So, possible values of X = 0, 1, 2,3.. Now, The chances of probability of hiring at least two candidates, P( X ≥ 2) = 60% = 0.60

The chances of probability that hiring of none of them = P( X = 0) = 10% = 0.10

The chances of probability that hiring of all of them = P( X = 3) = 5% = 0.05

We have to determine probability that the firm will make at least one hire, P(X ≥ 1).

By probability law, sum of any possible probability values = 1

=> P( X≥ 1 ) = 1 - P( X = 0)

= 1 - 0.10 = 0.90

b) The expected value, E(X), or we say mean μ of a discrete random variable X, is equals to the sum of resultant of multiply each value of the random variable by its probability. The formula is, E ( X ) = μ = ∑ x P ( x )

So, Probability that hiring of exactly two candidates, P( X= 2). As we know, P( X ≥ 2) = 0.60

=> P ( 3) + P (2) = 0.60

=> P(2) = 0.60 - 0.05 = 0.55

Probability that hiring of exactly one candidates, P( X= 1). From, P( X≥ 1 ) = 0.90

=> P( 1) + P ( 3) + P (2) = 0.90

=> P(1) = 0.30

Hence, excepted value, E(X) = ∑ x P ( x )

= 0 × 0.10 + 1× 0.30 + 2× 0.55 + 3× 0.05

= 1.57

Now, standard deviations, σ =√E(X²) - (E(X))²

E(X²) = ∑ x² P ( x )

= 0² ×0.10 + 1² ×0.30 + 2²× 0.55 + 3²× 0.05

= 3.07

so, the standard deviation of the number of hires = √3.07² - 1.57² = 0.78

Hence, required value is 0.78.

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