Your classroom has an area of 72 square feet wide. What is the perimeter of your classroom

Answer: n squared plus 10 n plus 25

Step-by-step explanation:

To make the expression a perfect square:

add a term that is equal to half the coefficient of n, squared.

the coefficient of n is 10, so half of it is 5

add 5 squared, or 25, to the expression:

n squared plus 10 n plus 25

this expression can be factored into (n+5) squared, which is a perfect square.

B. If the right-hand-side for the Labor Time constraint decreases to 400, the dual value (shadow price) will become a less negative number, such as -4.

This is because a decrease in the available resource (Labor Time) will generally cause the shadow price to move toward a less negative value, reflecting the increased scarcity of that resource in the cost minimization problem. The correct answer is D. If the right-hand-side for the Labor Time constraint decreases to 400, it means that there is less availability of labor time, which will increase the cost of the problem. As a result, the dual value (shadow price) will become more negative, such as -8, indicating that an additional unit of labor time constraint would now cost more to relax. The allowable increase in the Labor Time constraint will decrease, while the allowable decrease will increase.

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In the given problem, we can stack the sweets in six stacks, each with 24 sweets. So, there will be 24 Kaju burfies in each stack.

To stack the sweets in the least area, we want to minimize the number of stacks. To do this, we need to find the greatest common divisor (GCD) of 48 and 72, which is 24.

Therefore, we need to stack the sweets in groups of 24.

We have a total of 48 Kaju burfies, so we need to divide them into groups of 24.

48 / 24 = 2

So, we can stack the Kaju burfies in two stacks of 24 each.

We also have 72 badam becafio, which we need to stack in groups of 24.

72 / 24 = 3

So, we can stack the badam becafio in three stacks of 24 each.

Thus, we can stack the sweets in six stacks, each with 24 sweets.

So, there will be 24 Kaju burfies in each stack.

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In the given problem, we can stack the sweets in six stacks, each with 24 sweets. So, there will be 24 Kaju burfies in each stack.

To stack the sweets in the least area, we want to minimize the number of stacks. To do this, we need to find the greatest common divisor (GCD) of 48 and 72, which is 24.

Therefore, we need to stack the sweets in groups of 24.

We have a total of 48 Kaju burfies, so we need to divide them into groups of 24.

48 / 24 = 2

So, we can stack the Kaju burfies in two stacks of 24 each.

We also have 72 badam becafio, which we need to stack in groups of 24.

72 / 24 = 3

So, we can stack the badam becafio in three stacks of 24 each.

Thus, we can stack the sweets in six stacks, each with 24 sweets.

So, there will be 24 Kaju burfies in each stack.

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Answer:

Step-by-step explanation:

two angles are adjacent if they have a common side and a common vertex.

basically two angles that share a line

Verticcal angles are angles that are opposite of each other. like in ur example of the triangle for question 3,

angle 3 and 4 are vertical angles

The explicit function for the sequence is: [tex]f(n) = 3(4)^(n-1)[/tex]

An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant.we have to know the first term (a), the number of terms(n), and the common difference (d) between consecutive terms

To find the explicit function for the given sequence, we need to determine the values of a and b in the equation f(n) = [tex]a(b)^(n-1[/tex]), given that f(1) = 3.

We can find the value of a by substituting n=1 into the equation:

f(1) =[tex]a(b)^(1-1)[/tex]= a

3 = a

So, we will substitute 3 for a in the equation f(n) = [tex]a(b)^(n-1).[/tex]

To find the value of b, we can use the fact that the ratio between consecutive terms in the sequence is constant. We can calculate this ratio by dividing any term by its preceding term.

The ratio between the second and first terms is:

12/3 = 4

The ratio between the third and second terms is:

48/12 = 4

The ratio between the fourth and third terms is:

192/48 = 4

The ratio between the fifth and fourth terms is:

768/192 = 4

Since the ratio is constant and equal to 4, we can write:

b = 4

Therefore, the explicit function for the sequence is:

f(n) = [tex]3(4)^(n-1)[/tex]

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The molecular geometry of IF₄+ and IO₂F₂- are both see-saw.

However, SOF₄, SF₄, and XeO₂F₂ have different geometries - trigonal bipyramidal, square planar, and square pyramidal respectively. Therefore, the correct answer is "All of the following are see-saw except molecular geometry."

This question is testing the understanding of molecular geometry and its relationship to the number of lone pairs and bonding pairs around the central atom.

See-saw geometry has four bonding pairs and one lone pair around the central atom, while the other three compounds have different arrangements.

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complete question:

The molecular geometry of which of the  following are see-saw.(molecular  Geometry)

IF4+1

IO2F2−1

SOF4

SF4

XeO2F2

Answer:

21

Step-by-step explanation:

x² =49

4x = -28

total 21

Answer:

0.83333333....

Step-by-step explanation:

First conversation 5/6 onto division which is 5 divided by 6 which is 8.3

The required measure of the angle is m∠DCQ = 51.5° for tangent to the circle. The correct answer is option D.

Firstly, find the measure of arc ABC

As we know that the inscribed angle is half the length of the arc.

So, m∠D=(1/2)[arc ABC]

Here, m∠D=88°

Substitute and solve for arc ABC:

88°=(1/2)[arc ABC]

176° = [arc ABC]

arc ABC=176°

Now, finding the measure of arc DC:

As per the property of the complete circle,

arc ABC + arc AD + arc DC = 360°

Substitute the given values,

176° + 81° + arc DC = 360°

arc DC = 360°- 257°

arc DC = 103°

Now, Find the measure of the angle DCQ:

As we know that the inscribed angle is half the length of the arc.

So, m∠DCQ=(1/2)[arc DC]

Substitute the value of arc DC = 103°,

m∠DCQ=(1/2)[103°] = 51.5°

Thus, the required measure of the angle is m∠DCQ = 51.5°.

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Answer:

The rocket's height is increasing on the interval 0<t<2.

The set of strings that satisfy the condition "l= ω∈0, 1* | ω has exactly one pair of consecutive zeros" can be constructed by considering all possible positions of the consecutive zeros and constructing the rest of the string accordingly.

The term "l= ω∈0, 1*" means that we are considering all strings (ω) made up of 0's and 1's of length "l" where "l" is unknown but can be any positive integer. The "|" symbol indicates a condition that must be satisfied by the string. In this case, we are looking for strings that have exactly one pair of consecutive zeros.

To find such strings, we can start by considering the possible positions of the consecutive zeros. They could appear in the first two positions, the last two positions, or somewhere in between.

If the consecutive zeros appear in the first two positions, then the rest of the string can be any combination of 0's and 1's. Similarly, if the consecutive zeros appear in the last two positions, then the rest of the string can also be any combination of 0's and 1's.

However, if the consecutive zeros appear somewhere in between, then the rest of the string must be carefully constructed to ensure that no additional pairs of consecutive zeros appear. For example, if the consecutive zeros appear in the third and fourth positions, then the rest of the string must contain only one more zero and the remaining digits must be 1's.

Therefore, the set of strings that satisfy the condition "l= ω∈0, 1* | ω has exactly one pair of consecutive zeros" can be constructed by considering all possible positions of the consecutive zeros and constructing the rest of the string accordingly.

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The degree of the polynomial is 3

Algebraic expressions are defined as expressions that are made up of terms, variables, constants, factors and coefficients.

These expressions are also made up of mathematical operations, such as;

Polynomials are algebraic expressions with a degree that is greater than Note that the height exponent is the same as the degrees.

From the information given, we have;

C+cd² +6d³

Degree = 3

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The value of the quantity after 6 weeks is 7694.5

Percentage Increase is the difference between the final value and the initial value, expressed in the form of a percentage.

How to determine this

When an initial value = 5500

Grows at a rate of 0.95%

i.e 0.95% of 5500 = 52.25, it grows 52.25 per day

What is the value of the quantity after 6 weeks

When 7 days = 1 week

6 weeks = x

x = 6 * 7 days

x = 42 days

If it grows 52.25 per day

let x represent the value of quantity in 42 days

When 52.25 = 1 day

x = 42 days

x = 42 * 52.25

x = 2194.5

Therefore the value of the quantity after 6 weeks

= 2194.5 + 5500

= 7694.5

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The maximum displacement current through the cross-sectional area is 1.77 A.

The maximum displacement current through a cross-sectional area can be found using the equation,

[tex]I =\ \in_0(d \phi_{E/dt} )[/tex]

where I is the displacement current, ε₀ is the electric constant [tex]8.85 \times 10^{-12} F/m[/tex], and [tex](d \phi_{E/dt} )[/tex] is the rate of change of the electric flux through the cross-sectional area.

The electric flux [tex]\phi_E[/tex] through a surface is given by:

[tex]\phi_E = \int E\times dA[/tex]

where E is the electric field and dA is the differential area vector.

For a uniform electric field perpendicular to the surface, the electric flux through the surface is simply:

[tex]\phi_E = E\times A[/tex]

where E is the magnitude of the electric field and A is the area of the surface.

In this case, the magnitude of the electric field is given by:

[tex]E = E_0\ Cos(\omega t)[/tex]

The maximum value of E is [tex]E_0 =20 V/m[/tex], which occurs when [tex]Cos(\omega t) =1[/tex]

The maximum electric flux through the cross-sectional area [tex]A = 0.60 m^2[/tex] is therefore:

[tex]\phi_E = E \times A = (20 V/m) \times (0.60 m^2) = 12 V[/tex]

To find the maximum displacement current, we need to differentiate the electric flux with respect to time:

[tex]d\phi_{E/dt} = -E_0\ \omega Sin(\omega t)[/tex]

The maximum value of sin(ωt) is 1, so the maximum value of [tex]d\phi_{E/dt}[/tex] is:

[tex]d\phi_{E/dt} = -E_0\ \omega Sin(\omega t) = -20V/m \times (1.0 \times 10^7 s^{-1}) \times 1 \\d\phi_{E/dt} = -2.0 \times 10^8 V/s[/tex]

Therefore, the maximum displacement current through the cross-sectional area is:

[tex]I =\ \in_0(d \phi_{E/dt} ) = (8.85 \times 10^{-12} F/m)\ \times (-2.0 \times 10^8 V/s) =-1.77A\\[/tex]

The negative sign indicates that the displacement current is flowing in the opposite direction to the electric field. However, the magnitude of the displacement current is always positive.

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The maximum displacement current through the cross-sectional area of this volume is approximately 1.06 milliamperes (mA), or 1.06 × 10⁻³A.

In mathematics, maximum displacement refers to the maximum deviation of a point or object from its equilibrium position, as it undergoes a displacement or vibration.

The maximum displacement current through a cross-sectional area A can be found using the equation:

I = ε₀ A (dE/dt)

where ε₀ is the permittivity of free space, A is the cross-sectional area, and dE/dt is the time rate of change of the electric field.

In this case, the electric field is given by E = E₀ cos(ωt), so we can find its time derivative as follows:

dE/dt = -E₀ω sin(ωt)

The maximum displacement current occurs when sin(ωt) is equal to 1, which corresponds to the maximum value of the time-varying electric field. At this point, the displacement current is:

I = ε₀ A (dE/dt) = ε₀ A (-E₀ω)

Substituting the given values, we get:

I = (8.85 × 10⁻¹²C²/Nm²)(0.60 m²)(-20 V/m)(1.0 × 10⁷ s⁻¹)

I ≈ -1.06 × 10⁻³ A

Note that the negative sign indicates that the displacement current is flowing in the opposite direction to the time-varying electric field.

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Tthe integral of the joint PDF over its support is equal to 1, we have: ∫∫ fX,Y(x,y) dx dy = 1 2c = 1 c = 1/2 Therefore, the constant c is 1/2.

To find the constant c in the joint PDF of two random variables X and Y, given by fX,Y(x,y) = c, we need to use the property that the double integral of the joint PDF over the entire support equals 1. In this case, the support is defined by 0 ≤ x ≤ y ≤ 2.

Step 1: Set up the double integral
∫∫fX,Y(x,y) dx dy = 1

Step 2: Substitute fX,Y(x,y) with the given value
∫∫c dx dy = 1

Step 3: Determine the limits of integration
For x: 0 to y
For y: 0 to 2

Step 4: Solve the double integral
∫(from 0 to 2) ∫(from 0 to y) c dx dy = 1

Step 5: Integrate with respect to x
∫(from 0 to 2) [cx] (from 0 to y) dy = 1
∫(from 0 to 2) cy dy = 1

Step 6: Integrate with respect to y
[c/2 * y^2] (from 0 to 2) = 1
c(2^2)/2 - c(0^2)/2 = 1
c(4)/2 = 1

Step 7: Solve for c
c(2) = 1
c = 1/2

So, the constant c in the joint PDF fX,Y(x,y) = c is 1/2.

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These calculations, it appears that the sequence of partial sums is convergent, as the values of Sn appear to approach a limit as n increases.

To calculate the sequence of partial sums, we need to add up the first n terms of the series for each n up to 8. The nth term of the series is given by:

an = 2n / (n^5 + 1)

Therefore, the sequence of partial sums is:

S1 = 2/2 = 1

S2 = 2/2 + 3/26 = 1.5874

S3 = 2/2 + 3/26 + 4/641 = 1.3867

S4 = 2/2 + 3/26 + 4/641 + 5/15626 = 1.2600

S5 = 2/2 + 3/26 + 4/641 + 5/15626 + 6/390625 = 1.7000

S6 = 2/2 + 3/26 + 4/641 + 5/15626 + 6/390625 + 7/9765626 = 1.1006

S7 = 2/2 + 3/26 + 4/641 + 5/15626 + 6/390625 + 7/9765626 + 8/244140626 = 1.0455

S8 = 2/2 + 3/26 + 4/641 + 5/15626 + 6/390625 + 7/9765626 + 8/244140626 + 9/6103515626 = 1.0127

From these calculations, it appears that the sequence of partial sums is convergent, as the values of Sn appear to approach a limit as n increases.

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The value of BD in the similar triangle is 4 units.

A right angle triangle is a triangle that has one of its angles as 90 degrees. The triangles are similar .

Therefore, let's use the ratios of the similar triangle to find the side BD.

Let

BD = x

Therefore,

x / 6 = 6 / (5 + x)

cross multiply

x(x + 5) = 6 × 6

x² + 5x = 36

x² + 5x - 36 = 0

x² - 4x + 9x - 36 = 0

x(x - 4) + 9(x - 4) = 0

(x + 9)(x - 4) = 0

x = -9 or 4

Therefore, x(BD) can only be positive.

Hence,

BD = 4

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Entonces, si el club de teatro vende todas las latas de limonada y botellas de agua, ¿cuánto dinero recaudará?

Si venden 360 latas de limonada a $2 cada una, recaudarán $720.

Si venden 360 botellas de agua a $1.50 cada una, recaudarán $540.

En total, recaudarán $720 + $540 = $1260.

Entonces, el club de teatro reunirá más de los $500 que necesitan para cubrir el costo del alquiler del vestuario.

Answer:

y = 3/4x - 14

Step-by-step explanation:

The equation is y = mx + b

m = the slope

b = y-intercept

m = 3/4

The Y-intercept is located at (0, -14)

So, the equation of the line is y = 3/4x - 14

Answer:

10 - 7 = 3

Step-by-step explanation:

We Have facts are:

7 + 3 = 10

10 - 3 = 7

3 + 7 = 10

We are missing

10 - 7 = 3

So, the answer is 10 - 7 = 3

The volume of the solid with equation y = 1/4x^2, x=2when rotated the volume is  π/2 cubic units.

To find the volume of the solid obtained by rotating the region bounded by y=1/4x^2 and x=2 about the x-axis, we can use the disk or washer method.

First, let's sketch the region and the solid. The region is bounded by y=1/4x^2 and x=2, and looks like a quarter of a parabola with its vertex at the origin and passing through (2,1). When we rotate this region about the x-axis, we get a solid that looks like a bowl with a flat bottom and a curved side.

To find the volume of this solid, we need to integrate the area of each disk or washer. Since the region is bounded by x=2, we can set up our integral as follows:

V = ∫[0,2] π(1/4x^2)^2 dx

This represents the sum of the volumes of all the disks or washers from x=0 to x=2. Simplifying the integral, we get:

V = π/16 ∫[0,2] x^3 dx
V = π/16 * [x^4/4] from 0 to 2
V = π/16 * (2^4/4 - 0)
V = π/2

Therefore, the volume of the solid obtained by rotating the region bounded by y=1/4x^2 and x=2 about the x-axis is π/2 cubic units.

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Answer: 2

Step-by-step explanation:

hi

The unit normal vector to the surface f(x,y,z)=0 at the point p(2,5,-27) is (-1/sqrt(27), 5/sqrt(27), 1/sqrt(27))

To find a unit normal vector to the surface f(x, y, z) = ln(x - 5y - z) at the point P(2, 5, -27), you'll first need to compute the gradient of the function, which represents the normal vector.

The gradient is given by (∂f/∂x, ∂f/∂y, ∂f/∂z). Let's compute the partial derivatives:

∂f/∂x = 1/(x - 5y - z)

∂f/∂y = -5/(x - 5y - z)

∂f/∂z = -1/(x - 5y - z)

Now, evaluate the gradient at the point P(2, 5, -27):

∇f(P) = (1/(2 - 5*5 + 27), -5/(2 - 5*5 + 27), -1/(2 - 5*5 + 27))

∇f(P) = (1/-4, 5/4, 1/4)

Now we'll normalize this vector to get the unit normal vector:

||∇f(P)|| = sqrt[tex]((-1/4)^2[/tex] + [tex](5/4)^2[/tex] + [tex](1/4)^2)[/tex] = sqrt(27/16)

Unit normal vector = ∇f(P)/||∇f(P)|| = (-1/4, 5/4, 1/4) / (sqrt(27/16))

Unit normal vector = (-1/sqrt(27), 5/sqrt(27), 1/sqrt(27))

So, the unit normal vector to the surface at the point P(2, 5, -27) is (-1/sqrt(27), 5/sqrt(27), 1/sqrt(27)).

The Question was Incomplete, Find the full content below :

Find a unit normal vector to the surface f(x,y,z)=0 at the point p(2,5,-27) for the function f ( x , y , z ) = ln ( x − 5 y − z )

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y = 2 + 6e^(-3t).

We can solve the given initial value problem using the method of separation of variables.

Separating the variables, we get:

dy/(y-2) = -3 dt

Integrating both sides, we get:

ln|y-2| = -3t + C

where C is the constant of integration.

Using the initial condition y(0) = 8, we have:

ln|8-2| = C

C = ln(6)

Substituting the value of C, we get:

ln|y-2| = -3t + ln(6)

ln|y-2| = ln(6) - 3t

Taking exponential on both sides, we get:

|y-2| = e^(ln(6)-3t)

|y-2| = 6e^(-3t)

y-2 = ±6e^(-3t)

If we take the positive sign, we get:

y = 2 + 6e^(-3t)

Using the initial condition y(0) = 8, we get:

8 = 2 + 6e^(0)

Simplifying, we get:

6 = 6

Therefore, the solution to the given initial value problem is: y = 2 + 6e^(-3t)

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The measure of angle YXZ is 9.

We are given that;

XZ= x+54

YZ= x+108

Now,

By the property of angle sum of circle

x+54+x+108=180

2x+162=180

Solving the equation

2x=180-162

2x=18

x=9

Therefore, by the angle property the answer will be 9.

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The measure of angle YXZ is 9.

We are given that;

XZ= x+54

YZ= x+108

Now,

By the property of angle sum of circle

x+54+x+108=180

2x+162=180

Solving the equation

2x=180-162

2x=18

x=9

Therefore, by the angle property the answer will be 9.

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Answer:

Leading coefficient: -22

Degree: 2

Step-by-step explanation:

The given polynomial is:

-10v-18+v²-23v²

solving like terms, we get

-22v² - 10v - 18

The leading coefficient is the coefficient of the term with the highest degree. In this case, the term with the highest degree is -22v² and its coefficient is -22. Therefore, the leading coefficient is -22.

The degree of a polynomial is the highest power of the variable in the polynomial. In this case, the highest power of v is 2, which is the degree of the polynomial. Therefore, the degree of the polynomial is 2.

The 95% confidence interval for the difference in means is [-0.98, 10.98], which includes 0.

To calculate the 95% confidence limits on the difference between the means (m₁ - m₂) and the difference between the standard deviations (d), we can use the following formulas:

SE(m₁ - m₂) = √[(s₁²/n₁) + (s₂²/n₂)]

where s₁ and s₂ are the sample standard deviations, n₁ and n₂ are the sample sizes, and SE represents the standard error.

95% confidence interval for (m₁ - m₂) = (x₁ - x₂) ± (t(α/2) * SE(m₁ - m₂))

where x₁ and x₂ are the sample means, t(α/2) is the t-value for the appropriate degrees of freedom and alpha level, and SE(m₁ - m₂) is the standard error.

SE(d) = √[((n₁ - 1)s₁² + (n₂ - 1)s₂²)/(n₁ + n₂ - 2)] * √[1/n₁ + 1/n₂]

where s₁ and s₂ are the sample standard deviations, n₁ and n₂ are the sample sizes, and SE represents the standard error.

95% confidence interval for d = (s₁²/s₂²) * [(n₁ + n₂ - 2)/(n₁ - 1)] * F(α/2)

where F(α/2) is the F-value for the appropriate degrees of freedom and alpha level.

Using the given data, we have:

Expectation good: n₁ = 9, x₁ = 18, s₁ = 4.38

Expectation poor: n₂ = 8, x₂ = 17.125, s₂ = 4.373

SE(m₁ - m₂) = √[(s₁²/n₁) + (s₂²/n₂)] = √[(4.38²/9) + (4.373²/8)] = 1.913

Degrees of freedom = n₁ + n₂ - 2 = 15

t(α/2) = t(0.025) = 2.131

95% confidence interval for (m₁ - m₂) = (18 - 17.125) ± (2.131 * 1.913) = (0.546, 1.429)

SE(d) = √[((n₁ - 1)s₁² + (n₂ - 1)s₂²)/(n₁ + n₂ - 2)] * √[1/n₁ + 1/n₂] = √[((8)(4.373²) + (9)(4.38²))/(17)] * √[1/8 + 1/9] = 1.322

Degrees of freedom numerator = n₁ - 1 = 8

Degrees of freedom denominator = n₂ - 1 = 7

F(α/2) = F(0.025) = 4.256

95% confidence interval for d = (4.38²/4.373²) * [(9 + 8 - 2)/(8)] * 4.256 = (0.754, 3.880)

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a. The value of k that makes f(y) a density function is 0

b. The probability that Y lies within 2 standard deviations of its mean is 0.948.

Probability is a measure of the likelihood or chance of an event occurring. It is a number between 0 and 1, with 0 representing an impossible event and 1 representing a certain event. The probability of an event is calculated by dividing the number of ways the event can occur by the total number of possible outcomes.

a) To find the value of k that makes f(y) a density function, we need to integrate the density function from 0 to infinity and set it equal to 1 (since the total area under the density function should be equal to 1 for it to be a valid probability density function):

[tex]\int\limits0^\infty, ky^3e^{(-y/2)} dy = 1[/tex]

Using integration by parts, we can evaluate this integral as:

[tex]\rm [-2ky^3e^{(-y/2)} - 12ky^2e^{(-y/2)} - 24kye^{(-y/2)} - 48k][/tex]

evaluated from 0 to infinity

To make sure that the integral converges, we need to set the coefficient of

[tex]\rm e^{(-y/2)}[/tex] to zero.

Therefore, we have:- 2k = 0 [tex]\geq[/tex] k = 0

This implies that the probability density function f(y) is not valid, which means that there is a mistake in the given probability density function.

b) To determine if Y has a chi-square distribution, we need to compare its density function to the general form of the chi-square distribution. The density function of the chi-square distribution with n degrees of freedom is:

[tex]\rm f(x) = (1/2^{(n/2)} \Gamma (n/2))x^{(n/2-1)}e^{(-x/2)}, x > 0[/tex]

where Γ is the gamma function.

Comparing this to the given density function, we see that it is not of the same form, so Y does not have a chi-square distribution.

To find the mean and standard deviation of Y, we can use the formulae:

Mean = E(Y) =

[tex]\rm \int\limits 0^\infty yf(y)dy[/tex]

Standard deviation = √(V(Y)) = √(E(Y²) - [E(Y)]²)

Using integration by parts, we can evaluate the mean as:

E(Y) = 6

To evaluate the expected value of Y², we can use integration by parts twice:

[tex]\rm E(Y^2) = \int\limits 0^\infty y^2 f(y)dy= 20[/tex]

Therefore, the standard deviation of Y is:

Standard deviation = √(E(Y²) - [E(Y)]²) = √(20 - 6²) = √(4) = 2d)

The probability that Y lies within 2 standard deviations of its mean can be calculated as:

P(mean - 2SD &lt; Y &lt; mean + 2SD) = P(6 - 22 [tex]<[/tex] Y [tex]<[/tex] 6 + 22) = P(2 [tex]<[/tex] Y [tex]<[/tex] 10)

Using the probability density function, we can evaluate this probability as:

[tex]\rm \int\limits 2^{10} ky^3e^{(-y/2)} dy[/tex]

This integral can be evaluated numerically or by using integration by parts. The result is approximately 0.948, hence, the probability that Y lies within 2 standard deviations of its mean is 0.948.

To know more about probability visit:

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