Tia types 66 words in 3 minutes. She wants to know how many words she can type in 5 and 10 minutes if she keeps the same rate.

The probability that randomly selected mechanical component weighs: more than 11.5 lbf is 0.0099, less than 8.7 lbf is 0.3208,  less than 5.0 lbf is 0.0228, between 6.2 lbf and 7.0 lbf is 0.1314, between 10.3 lbf and 14.0 lbf is 0.0436, between 6.8 lbf and 8.9 lbf is 0.3464.

We can use the standard normal distribution and z-scores to answer these questions:

P(X > 11.5) = P(Z > (11.5 - 8) / 1.5) = P(Z > 2.33) = 0.0099

Therefore, the probability that a randomly selected mechanical component weighs more than 11.5 lbf is 0.0099, or about 1%.

P(X < 8.7) = P(Z < (8.7 - 8) / 1.5) = P(Z < 0.47) = 0.3208

Therefore, the probability that a randomly selected mechanical component weighs less than 8.7 lbf is 0.3208, or about 32%.

P(X < 5.0) = P(Z < (5 - 8) / 1.5) = P(Z < -2) = 0.0228

Therefore, the probability that a randomly selected mechanical component weighs less than 5.0 lbf is 0.0228, or about 2%.

P(6.2 < X < 7.0) = P((6.2 - 8) / 1.5 < Z < (7 - 8) / 1.5) = P(-1.2 < Z < -0.67) = 0.1314

Therefore, the probability that a randomly selected mechanical component weighs between 6.2 lbf and 7.0 lbf is 0.1314, or about 13%.

P(10.3 < X < 14.0) = P((10.3 - 8) / 1.5 < Z < (14 - 8) / 1.5) = P(1.53 < Z < 2.67) = 0.0436

Therefore, the probability that a randomly selected mechanical component weighs between 10.3 lbf and 14.0 lbf is 0.0436, or about 4%.

P(6.8 < X < 8.9) = P((6.8 - 8) / 1.5 < Z < (8.9 - 8) / 1.5) = P(-0.47 < Z < 0.60) = 0.3464

Therefore, the probability that a randomly selected mechanical component weighs between 6.8 lbf and 8.9 lbf is 0.3464, or about 35%.

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____The given question is incomplete, the complete question is given below:

Suppose that the weight of a mechanical component is normally distributed with mean p = 8.0 Ibf and standard deviation = 1.5 lbf. Answer the following questions: 1. If you randomly select a mechanical component, what is the probability that it weighs more than 11.5 lbf? 2. If you randomly select a mechanical component, what is the probability that it weighs less than 8.7 lbf? 3. If you randomly select a mechanical component, what is the probability that it weighs less than 5.0 lbf? 5. If you randomly select a mechanical component, what is the probability that it weighs between 6.2 lbf and 7.0 lbf? 6. If you randomly select a mechanical component, what is the probability that it weighs between 10.3 lbf and 14.0 lbf? 7. If you randomly select a mechanical component, what is the probability that it weighs between 6.8 lbf and 8.9 lbf?

A data of cases of diabetes from the centers for disease control and prevention, the probability that the combined sample tests positive with at least 1 of the 10 people having diabetes is equals to 0.04218.

We have a data from the centers for disease control and prevention.

The rate of new cases of diabetes

= 4.3 per 1000

So, probability ( diabetes) =\( \frac{4.3}{1000}\) = 0.0043

Using complement rule, Probability for no diabetes people, P( no diabetes)

= 1 - 0.0043 = 0.9957

Now, blood samples from 10 is randomly selected. It is assumed that each of these different people having diabetes is independent events. Using multiplcation rule for independent events, P( All 10 have no diabetes )

= P( no diabetes)× P( no diabetes)×....× P( no diabetes) ( 10 times)

= ( P( no diabetes))¹⁰ = 0.9957¹⁰

= 0.957823

Using complement rule, P ( atleast 1 the 10 people having diabetes) = 1 - P( All 10 have no diabetes ) = 1 - 0.957823

= 0.04218

Since, probability value is small so it is unlikely that a combined sample test. Hence, required probability is 0.04218.

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Although part of your question is missing, you might be referring to this full question: If the 1st person saves 1/4 of total, 2nd person saves 2/3 of total, and 3rd person saves 1/10, which fraction left to pay for the birthday party?

The fraction left to pay for the birthday party is 17/30.

The calculation is as follows:

1 * 1/4 = 1/4 … (1)

1/4 * 2/3 = 2/12 … (2)

2/12 * 1/10 = 2/120 … (3)

Fraction left to pay:

= 1 - (1/4 + 2/12 + 2/120)

= 1 - (30/120 + 20/120 + 2/120)

= 1 - (52/120)

= 1 - 13/30

= 30/30 - 13/30

= 17/30

Thus, the fraction left to pay for the birthday party is 17/30.

A fraction is a part of a whole. In arithmetic, the number is expressed as a quotient, where the numerator is divided by the denominator. In a simple fraction, both are integers. A complex fraction has a fraction in the numerator or denominator. In a proper fraction, the numerator is less than the denominator.

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Based on the information given, it appears that the consumer rejected the null hypothesis that the mean price for a gallon of unleaded gasoline in the United States in a certain year is $3.110. This means that the consumer believed the mean price to be different from $3.110. However, in reality, the mean price for a gallon of unleaded gasoline in the United States in that year was $3.210.

Yes, an error was made in this situation. The consumer's hypothesis test was aimed at determining whether the mean price for a gallon of unleaded gasoline in the United States in a certain year is different from $3.110. Since the consumer rejected the null hypothesis, they concluded that the mean price is indeed different from $3.110.

Therefore, the consumer made a type I error. A type I error is made when a null hypothesis is rejected when it is actually true. In this case, the consumer rejected the null hypothesis that the mean price for a gallon of unleaded gasoline in the United States in a certain year is $3.110, when in fact, the true mean price was $3.210.

In reality, the mean price for a gallon of unleaded gasoline in the United States in that year is $3.210. Since the true mean is different from the hypothesized mean of $3.110, the consumer's decision to reject the null hypothesis was correct.

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Answer:

65%

Step-by-step explanation:

Hope this helps! Pls give brainliest!

Answer:

53%

Step-by-step explanation:

In summary, the girls' data has a larger spread in terms of variance and standard deviation, while the boys' data has a larger range.

In mathematics, the mean is a measure of central tendency of a set of numbers, which is also known as the average. It is obtained by adding up all the numbers in the set and then dividing the sum by the total number of items in the set. The mean is often used to describe the typical value of a dataset.

Here,

To compute the measures of spread for the data collected for boys and girls, we need to calculate the range, interquartile range (IQR), and standard deviation.

For the girls:

Range = 81 - 15 = 66

IQR = Q3 - Q1 = 56 - 32 = 24

Standard deviation = 23.96

For the boys:

Range = 81 - 0 = 81

IQR = Q3 - Q1 = 45 - 22 = 23

Standard deviation = 26.93

The range measures the difference between the largest and smallest values in the data set. In this case, the range for girls is smaller than the range for boys, indicating that the data for girls is less spread out than the data for boys.

The interquartile range (IQR) measures the spread of the middle 50% of the data. The IQR for girls is smaller than the IQR for boys, again indicating that the data for girls is less spread out than the data for boys.

The standard deviation measures the average deviation of the data from the mean. The standard deviation for boys is larger than the standard deviation for girls, indicating that the data for boys is more spread out than the data for girls.

b) To compute the measures of spread, we need to find the range, variance, and standard deviation for both the boys and girls data.

For the girls data:

Range = 81 - 15 = 66

Variance = 4143.3

Standard deviation = 64.36

For the boys data:

Range = 81 - 0 = 81

Variance = 947.9

Standard deviation = 30.82

The range for the boys data is larger than the range for the girls data, indicating that the boys' scores are more spread out than the girls' scores. However, when we look at the variance and standard deviation, we see that the girls' data has a much larger spread than the boys' data. This means that the girls' scores are more varied and spread out from the mean than the boys' scores.

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The partial derivative diffusivity equation for spherical flow is ∂²p/∂r² + (1/r) ∂p/∂r = ϕμC/k ∂p/∂t, and for hemispherical flow, it is the same equation.

1. The partial derivative diffusivity equation for spherical flow is derived from the spherical coordinate system and applies to radial flow in a spherical geometry. It can be expressed as ∂²p/∂r² + (1/r) ∂p/∂r = ϕμC/k ∂p/∂t.

2. The partial derivative diffusivity equation for hemispherical flow is derived from the hemispherical coordinate system and applies to radial flow in a hemispherical geometry. It can be expressed as ∂²p/∂r² + (1/r) ∂p/∂r = ϕμC/k ∂p/∂t.

1. For the derivation of the partial derivative diffusivity equation for spherical flow, we consider a spherical coordinate system with the radial direction (r), the azimuthal angle (θ), and the polar angle (φ). By assuming steady-state flow and neglecting the other coordinate directions, we focus on radial flow. Applying the Laplace operator (∇²) in spherical coordinates, we obtain ∇²p = (1/r²) (∂/∂r) (r² ∂p/∂r). Simplifying this expression, we arrive at ∂²p/∂r² + (1/r) ∂p/∂r.

2. Similarly, for the derivation of the partial derivative diffusivity equation for hemispherical flow, we consider a hemispherical coordinate system with the radial direction (r), the azimuthal angle (θ), and the elevation angle (ε). Again, assuming steady-state flow and neglecting the other coordinate directions, we focus on radial flow. Applying the Laplace operator (∇²) in hemispherical coordinates, we obtain ∇²p = (1/r²) (∂/∂r) (r² ∂p/∂r). Simplifying this expression, we arrive at ∂²p/∂r² + (1/r) ∂p/∂r.

In both cases, the term ϕμC/k ∂p/∂t represents the source or sink term, where ϕ is the porosity, μ is the fluid viscosity, C is the compressibility, k is the permeability, and ∂p/∂t is the change in pressure over time.

These equations are commonly used in fluid mechanics and petroleum engineering to describe radial flow behavior in spherical and hemispherical geometries, respectively.

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Have a great day

Answer:

7y-3x-8z

Step-by-step explanation:

(3y+4y)+(2x-5x)-8z

7y+(2x-5x)-8z

7y-3x-8z

The first-order perturbation contribution to the energy eigenvalues is ΔEₙ(1) = ab/4.

To calculate the first-order perturbation contributions to the energy eigenvalues, we need to apply the time-independent perturbation theory. Given that the perturbation term is H' = bx over the interval 0 ≤ x ≤ a, we can write the perturbed Hamiltonian as:

H = H₀ + H' = H₀ + bx

where H₀ is the unperturbed Hamiltonian, given by H₀ = -ħ²/2m * d²/dx² in this case. We know the unperturbed energy eigenvalues are given by Eₙ = (ħ²π²/2ma²) * n², where n = 1, 2, 3, ... To find the first-order perturbation correction to the energy eigenvalues, we use the formula:

ΔEₙ(1) = ⟨ψₙ|H'|ψₙ⟩

where ⟨ψₙ|H'|ψₙ⟩ is the expectation value of the perturbation Hamiltonian H' in the nth energy eigenstate ψₙ. Let's calculate it step by step: Express the energy eigenstate ψₙ(x) in terms of a:

ψₙ(x) = (a/√2) * sin(nπx/a)

Calculate the expectation value ⟨ψₙ|H'|ψₙ⟩:

⟨ψₙ|H'|ψₙ⟩ = ∫₀ᵃ (a/√2) * sin(nπx/a) * bx * (a/√2) * sin(nπx/a) dx

= ab/(2) * ∫₀ᵃ sin²(nπx/a) dx

Apply the trigonometric identity sin²θ = (1 - cos(2θ))/2:

⟨ψₙ|H'|ψₙ⟩ = ab/(2) * ∫₀ᵃ (1 - cos(2nπx/a))/2 dx

= ab/(4) * ∫₀ᵃ (1 - cos(2nπx/a)) dx

= ab/(4) * (x - (a/2nπ) * sin(2nπx/a)) ∣₀ᵃ

= ab/(4) * (a - (a/2nπ) * sin(2nπ) - 0)

Note: Since sin(2nπ) = sin(0) = 0, the second term vanishes.

⟨ψₙ|H'|ψₙ⟩ = ab/(4) * a

= ab/4

Substitute ⟨ψₙ|H'|ψₙ⟩ into the formula for the first-order perturbation correction:

ΔEₙ(1) = ⟨ψₙ|H'|ψₙ⟩

= ab/4

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Absolute value which means even negative integers are considered as positive

Answer:Therefore, the area of the circle with a radius of 27 mm is approximately 2286.2 mm².

Step-by-step explanation:Area = π * r^2

Where:

π (pi) is approximately 3.14159

r is the radius of the circle

In this case, the given radius is 27 mm. Plugging the value into the formula, we have:

Area = 3.14159 * (27 mm)^2

Calculating this:

Area ≈ 3.14159 * (27 mm * 27 mm)

Area ≈ 3.14159 * 729 mm²

Area ≈ 2286.19359 mm²

Rounding the area to one decimal place:

The duration of this bond is 6.5 years.

Duration is a measure of a bond's sensitivity to changes in interest rates. To calculate the duration of this bond, we need to consider the present value of the bond's cash flows and the timing of those cash flows. In this case, the bond has a face value of $1,000, a coupon rate of 10 percent, and annual coupon payments.

First, we calculate the present value of the bond's cash flows. Since the bond has a coupon rate of 10 percent, the annual coupon payment is $100 ($1,000 x 10%). The bond has a remaining maturity of seven years, so there will be seven coupon payments in total. We can calculate the present value of these cash flows using the formula for the present value of an ordinary annuity:

Present Value = Coupon Payment x [1 - (1 + interest rate)^(-number of periods)] / interest rate

Assuming an interest rate of r, we have:

Present Value = $100 x [1 - (1 + r)^(-7)] / r

Next, we need to find the yield to maturity (YTM) of the bond. YTM is the rate of return an investor would earn by holding the bond until maturity. Since the bond is currently selling for $1,104 in the market, we can set up the following equation:

$1,104 = Present Value + (Coupon Payment / (1 + r)^7)

By solving this equation for r, we can find the yield to maturity. Using a financial calculator or spreadsheet software, we can determine that the yield to maturity is approximately 7 percent.

Now, we can calculate the duration of the bond. The duration formula is the weighted average time until the bond's cash flows are received, where the weights are the present values of the cash flows. In this case, we have seven annual cash flows, so the duration can be calculated as follows:

Duration = [(1 x Present Value of Year 1) + (2 x Present Value of Year 2) + ... + (n x Present Value of Year n)] / Present Value of the Bond

Plugging in the values, we get:

Duration = [(1 x Present Value of Year 1) + (2 x Present Value of Year 2) + ... + (7 x Present Value of Year 7)] / Present Value of the Bond

Calculating the present values for each year using an interest rate of 7 percent, we find:

Present Value of Year 1 = $100 / (1 + 0.07)^1

Present Value of Year 2 = $100 / (1 + 0.07)^2

...

Present Value of Year 7 = $100 / (1 + 0.07)^7

After calculating the present values for each year and plugging them into the formula, we find that the duration of the bond is approximately 6.5 years.

Therefore, the correct answer is 6.5 years.

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it may be permissible to proceed with a hypothesis test even if the assumption of random participant selection is violated, under the conditions of known and accounted for non-random selection or random assignment to treatment groups

Random participant selection is an important assumption in hypothesis testing, as it helps ensure the generalizability of the results to the target population. However, in some situations, it may be impractical or impossible to achieve perfect random selection. In such cases, there are a few conditions under which it may still be permissible to proceed with a hypothesis test despite the violation of this assumption:

Non-random selection is known and accounted for: If the non-random selection process is well-documented and understood, researchers can adjust their analysis or statistical methods to account for potential biases introduced by the non-random selection.

Random assignment to treatment groups: Even if participants are not randomly selected, random assignment to different treatment groups can help mitigate the impact of non-random selection. By randomly assigning participants to treatment groups, the effects of non-random selection are distributed evenly across the groups, allowing for valid comparisons and hypothesis testing.

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Using missing length of each triangle are-

Part 6 (a) c = 4.25 m.

Part 6 (b) c = 1.65 m

The formula for the cosine rule:

cosФ = (a² + b² - c²)/2ab

Ф is the angle between sides a and b.

c is the side just opposite to the angle Ф.

The given values in question.

Part 6 (a)

a = 4.4m, b = 3.2m, Ф = 66°.

cos 66 = (4.4² + 3.2² - c²)/2(3.2)(4.4)

On simplification;

c² = 18.24

c = 4.25 m

(b) a = 1.8m, b = 2.2m, Ф = 46°.

cos 46 = (1.8² + 2.2² - c²)/2(1.8)(2.2)

On simplification;

c² = 2.72

c = 1.65 m

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The correct question is  question 6. (a) and (b).

Answer:

C

Step-by-step explanation:

the equation of the tangent line to the graph of the function at the point (0, 0) is y = 5x.

To find the equation of the tangent line to the graph of the function at the point (0, 0), we need to find the derivative of the function and evaluate it at x = 0.

The given function is y = 5 ln(exe^(-x^2)).

To find the derivative, we can use the chain rule and the properties of logarithmic and exponential functions. The derivative of y with respect to x can be calculated as follows:

dy/dx = 5 * (1/exe^(-x^2)) * (d/dx(exe^(-x^2)))

Applying the chain rule, we have:

dy/dx = 5 * (1/exe^(-x^2)) * (e^(-x^2) * d/dx(ex) + ex * d/dx(e^(-x^2)))

Simplifying further, we get:

dy/dx = 5 * (1/exe^(-x^2)) * (e^(-x^2) * 1 + ex * (-2x))

dy/dx = 5 * (e^(-x^2) - 2xex) / (ex * e^(-x^2))

Now, we can evaluate the derivative at x = 0 to find the slope of the tangent line at the point (0, 0).

dy/dx = 5 * (e^0 - 2(0)e^0) / (e^0 * e^0) = 5 * (1 - 0) / 1 = 5

Therefore, the slope of the tangent line at the point (0, 0) is 5.

Using the point-slope form of a linear equation (y - y1 = m(x - x1)), where m is the slope and (x1, y1) is the given point, we can substitute the values to find the equation of the tangent line:

y - 0 = 5(x - 0)

Simplifying, we get:

y = 5x

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Answer:

24

Step-by-step explanation:

Answer:

Mean = 260.45

Median = 284

Mode = 287, 285  (there are 2 modes)

Since it is a championship sports team , it is likely that this data is representative of all players in that league

Step-by-step explanation:

Mean = Sum of the weights/11 = 2865/11 = 260.45

Median is the middle of the data set when the data values are arranged in ascending order

In ascending order: 206, 216, 217, 235, 273, 284, 285, 285, 287, 287, 290

Mode is the value that occurs the maximum number of times in a data set

Here there are 2 modes 285 and 287 both of which occur twice

A firm experiences increasing returns to scale if inputs are doubled and output more than doubles.

When the firm's output grows at a faster rate than the growth in inputs, increasing returns to scale result. In this case, the company experiences economies of scale, which makes it more effective as it grows its production.

The firm is able to boost productivity and efficiency as it expands its scale of operations if inputs are doubled and output more than doubles.

This can be ascribed to a number of things, including specialisation, labour division, the use of capital-intensive technology, discounts for bulk purchases, and spreading fixed costs over a higher output. Lower average costs per unit of output result in higher profitability and competitiveness for the company.

The firm gains a number of benefits from growing returns to scale. First off, it lets the company to benefit from cost savings brought about by economies of scale, allowing it to manufacture goods or services for less money per unit. This may enable more competitive pricing on the market or result in larger profit margins.

Second, raising returns to scale can result in better operational effectiveness and resource utilisation. As the company grows in size, it will be able to use resources more wisely and profit from production volume-related synergies.market prices that are competitive.

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The resultant function is: L(1.1) ≈ 16√2 + 0.4`.

Given the function `f(x)=√128x³ + 384`.

The first derivative of `f(x)` is `f′(x) = 96x²/√128x³ + 384`

Let `x = 1` in `f′(x)` to obtain the slope of the tangent line at `x = 1`f′(1)

= `96(1)²/√128(1³) + 384`

= `96/24`

= `4`

The point at `x = 1` is `(1,f(1))`.

To find `f(1)`, substitute `x = 1` into the original function.f(1) = `√128(1³) + 384` = `√512` = `16√2`

Using the point-slope form of the equation of a line with slope `m = 4` and passing through `(1,16√2)` yields:

L(x) = `4(x - 1) + 16√2`

L(x) = `4x - 4 + 16√2`

L(x) = `4x + 16√2 - 4`

The equation of the tangent line is `L(x) = 4x + 16√2 - 4`.

To approximate `f(1.1)` using `L(x)`, substitute `x = 1.1`.

L(1.1) = `4(1.1) + 16√2 - 4`

L(1.1) = `4.4 + 16√2 - 4`

L(1.1) = `16√2 + 0.4`

Thus, `L(1.1) ≈ 16√2 + 0.4`.

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Answer:

4 gallons

Step-by-step explanation:

We can write a ratio to solve

95 miles             76 miles

----------------- = -------------

5 gallons            x gallons

Using cross products

95 * x = 5 * 76

Divide each side by 95

x = 5 * 76 / 85

x =4

To find the integral ∫∫c y^2 dx + x dy over the triangle with vertices (0, 0), (1, 1), and (0, 1), assuming counterclockwise orientation, we can evaluate the double integral by dividing the triangle into two smaller regions and integrating over each region separately.

The given integral can be split into two separate integrals representing the two regions of the triangle. We can divide the triangle into two triangles, one with vertices (0, 0), (1, 0), and (1, 1), and the other with vertices (0, 0), (0, 1), and (1, 1).

For the first triangle, the limits of integration are:

x: from 0 to 1

y: from 0 to x

The integral over this region becomes:

∫∫c y^2 dx + x dy = ∫[0,1] ∫[0,x] y^2 dx + x d

Evaluating this integral will give us the contribution of the first triangle to the overall result.

For the second triangle, the limits of integration are:

x: from 0 to 1

y: from x to 1

The integral over this region becomes:

∫∫c y^2 dx + x dy = ∫[0,1] ∫[x,1] y^2 dx + x dy

Evaluating this integral will give us the contribution of the second triangle to the overall result.

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Answer: B) 112 m

Step-by-step explanation: Hope this helps! There are 168 hours in a week. You divide by 3 because it moves 2 meters every 3 hours. Then you multiply by 2.

Answer:

x =65

Step-by-step explanation:

You see it is very hard

2 S V II T U

3 S T II V U

7 Definition of parallel lines

8 A parallelogram has two sets of parallel lines. The shape in the picture has two sets of parallel lines, therefore the shape is a parallelogram

To compute the Laplace transform of the given function, we can use the linearity property of the Laplace transform and apply the transform to each term separately.

Using the Laplace transform pairs:

L{1} = 1/s

L{u(t)} = 1/(s+1)

L{e^(-6t)} = 1/(s+6)

L{cos(πt)} = s/(s^2+π^2)

Applying these transforms to the given function:

L{1 u^(5/2)(t) e^(-6t) cos(πt)} = L{1} * L{u^(5/2)(t)} * L{e^(-6t)} * L{cos(πt)}

Substituting the transform pairs:

= (1/s) * (1/(s+1)^(5/2)) * (1/(s+6)) * (s/(s^2+π^2))

Simplifying this expression, we can multiply the terms together:

= s / (s(s+1)^(5/2)(s+6)(s^2+π^2))

Therefore, the Laplace transform of the given function is:

L{1 u^(5/2)(t) e^(-6t) cos(πt)} = s / (s(s+1)^(5/2)(s+6)(s^2+π^2))

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Answer:

Step-by-step explanation:

\((\frac{13 - \sqrt{57} }{8} , \frac{13 + \sqrt{57} }{8} )\)

Answer:

if use celebrates help to advertise cosmetic product than it is nothing new.

if we create cosmetic limit edition advertise than we will watch the profit grow high

if a friend of hers can advertise cosmetic products on her site than we could advertise them in our brochure

if we use a communication channels to advertise our cosmetic product than we might make it popular enough to sell them a bunch.

(4 samples)

I hope it helps.

Answer:

1. If you use this lotion, then your skin will glow

2. If you put this makeup in your face, then you will look stunning.

3. If you use this lip gloss, then your lip will not become dry.

4. If you use this soap, then germs in your body will remove.

5. If you use this hair product, then your hair will become bouncy.

Step-by-step explanation:

Hope it helps:)

Answer:

≤

Step-by-step explanation:

First we solve 5(x + 2) ≥ 2(x + 8).

5x + 10 ≥ 2x + 16

3x ≥ 6

x ≥ 2.

Now we place an equal sign into the box:

4(x - 6) = 2(x - 10)

4x - 24 = 2x - 20

2x = 4

x = 2.

For the solution to only have 1 value, the solution must be x = 2.

Therefore we must have x ≥ 2 and x ≤ 2.