A student with a poor understanding of straight-line graphs writes down some incorrect information
next to each equation. Decide how the error might have been made and then correct the information.

X-2y=4. (Gradient=1)

Answer:

21

Step-by-step explanation:

5. (1+ 3.2)

5 x 1 = 5

5 x 3.2 = 16

16 +5  = 21.

= 21

Answer:

4.2

Step-by-step explanation:

1 + 3.2= 4.2

that is all u have have as explanation!

Answer:

C. each pillow costs 9.5$ each.

The student covers a total distance of 22.05 kilometers in a week's time, walking between the park and the house each day.

To find the total distance covered by the student in a week's time, we need to calculate the distance covered in one round trip (from the house to the park and back) and then multiply it by the number of round trips in a week.

Given that the difference between the park and house is 1 kilometer and 575 meters, we can convert it to a total distance of 1.575 kilometers.

In a round trip, the student covers twice the distance between the park and the house, which is 1.575 kilometers * 2 = 3.15 kilometers.

Now, we need to determine how many round trips the student makes in a week. Let's assume the student makes one round trip each day.

Since there are 7 days in a week, the total distance covered by the student in a week's time is 3.15 kilometers * 7 = 22.05 kilometers.

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Nikola took 2 hours to finish the swimming part

Nikola took 4.5 hours to finish the running part

A form of arithmetic in which the rules of arithmetic are used to combine letters that represent numbers.

Calculations:

Assume,

\(t=time\;in\;hrs\;to\;complete\;swimming\;part\)

\(6.5-t=time\;in\;hrs\;to\;complete\;running\;part\)

\(d=distance\;in\;km\;that\;he\;swam\)

\(73.5-d=distance\;in\;km\;that\;he\;ran\)

Equation of swimming part,

\(d=3t\)

Equation of running part.

\(73.5-d=15\times(6.5-t)\)

Substitute the value of d in running part equation,

\(73.5-3t=15\times(6.5-t)\)

\(73.5-3t=97.5-15t\)

\(12t=24\)

Therefore,

\(t=2\)

\(Time\;to\;complete\;summing\;part=t\)

\(Time\;to\;complete\;summing\;part=2\;hours\)

\(Time\;to\;complete\;summing\;part=6.5-t\)

\(Time\;to\;complete\;summing\;part=6.5-2\)

\(Time\;to\;complete\;summing\;part=4.5\;hours\)

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Answer: A=9x^4

Step-by-step explanation:

Hii, do you need to find the area of a square with a side length of 3x^2? Let me help you out! (:

To find the area of a square, you need to square its side length.

Sounds strange, right?

The formula is shown below.

\(\LARGE\boldsymbol{A=a^2}\)

Here "A' (capital A) denotes the area

"a" (lower-case a) denotes the side length

The 2 above the a tells us that we need to multiply the side length times itself.

Alright, now it's just a matter of sticking in the known values...

\(\sf A=(3x^2)^2\)

Simplify

\(\sf A=9x^4\)

Voila! There's our answer, cheers! (;

Hope I helped! Best wishes! (:

Reach far. Aim high. Dream big.

\(\underbrace\)

To calculate the minimum sample size needed, we can use the formula: n = (z * σ / E)^2

Where:

z = the z-score associated with the desired level of confidence (let's assume 95% confidence, so z = 1.96)

σ = the standard deviation of the population (given in the problem statement)

E = the maximum error margin (given in the problem statement)

Plugging in the values, we get:

n = (1.96 * σ / E)^2

To determine the value of σ/E, we need more information. Let's assume that the maximum error margin E is 0.5, which means that the psychologist wants to be within 0.5 points of the true population mean si score.

Now, let's say that σ = 10 (just as an example). Then, σ/E = 10/0.5 = 20.

Plugging this into the formula, we get:

n = (1.96 * 10 / 0.5)^2 = 384.16

Rounding up to the nearest whole number, the minimum sample size needed is 385.

Therefore, the psychologist would need to take a random sample of at least 385 si scores from smokers to be confident that her estimate of the population mean si score is within 0.5 points.

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A bin of 50 parts contains 5 that are defective and a sample of 10 parts is selected without replacement, then total 40,753,713 samples contain at least 4 defective parts.

A bin of 50 parts contains 5 that are defective.

A sample of 10 parts is selected without replacement.

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = Samples with precisely 4 defectives + samples with precisely 5 defects

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = \(^{5}C_{4}\cdot ^{45}C_{6}+ ^{5}C_{5}\cdot^{45}C_{5}\)

Using the formula \(^nC_{r} = \frac{n!}{r!(n-r)!}\)

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = \(\frac{5!}{4!(5-4)!}\cdot\frac{45!}{6!(45-6)!}+\frac{5!}{5!(5-5)!}\cdot\frac{45!}{5!(45-5)!}\)

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = \(\frac{5!}{4!1!}\cdot\frac{45!}{6!39!}+\frac{5!}{5!0!}\cdot\frac{45!}{5!40!}\)

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = \(\frac{5\times4!}{4!\times1}\cdot\frac{45\times44\times43\times42\times41\times40\times39!}{6\times5\times4\times3\times2\times1\times39!}+1\cdot\frac{45\times44\times43\times42\times41\times40!}{5\times4\times3\times2\times1\times40!}\)

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = 5 × 15 × 44 × 43 × 7 × 41 + 1 × 9 × 11 × 7 × 41

Number of samples of size 10 from 50 pieces, at least 4 of which are faulty = 40,753,713

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For vectors \(\(\overrightarrow a\)\) and \(\(\overrightarrow b\)\) , the cross product is given as:

\(\[\overrightarrow a \times \overrightarrow b= \begin{vmatrix}\ i & j & k \\a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \end{vmatrix}\]\)

where i, j, k are the unit vectors, a1, a2, a3, b1, b2, b3 are the components of the vectors \(\(\overrightarrow a\) and \(\overrightarrow b\)\)respectively.

Question 1: Given the vectors

\(\( a=i-2 j+3 \boldsymbol{k} \) and \( \boldsymbol{b}=-2 \boldsymbol{i}+3 \boldsymbol{j}-\boldsymbol{k} \\)).

Find a.\(\( \boldsymbol{a} \times \boldsymbol{b}\)

Given that vectors \(\(\overrightarrow a= i-2j+3k \)and \(\overrightarrow b= -2i+3j-k \)\)

We are to find the vector a × b Using the cross product formula,

we get,

\(\[\overrightarrow a \times \overrightarrow b= \begin{vmatrix}\ i & j & k \\1 & -2 & 3 \\-2 & 3 & -1 \end{vmatrix}\]\)

Evaluating the determinant, we get\(\[\begin{aligned}\ \overrightarrow a \times \overrightarrow b & = \left(i\left(3\right) -j\left(-1\right) +k\left(9\right)\right)\mathbf{i} -\left(i\left(-2\right) -j\left(-2\right) +k\left(-2\right)\right)\mathbf{j} \\&\quad +\left(i\left(3\right) -j\left(-4\right) +k\left(-5\right)\right)\mathbf{k} \\& = \mathbf{3i+6j-15k}\end{aligned}\]\)

We are given two vectors \(\(\overrightarrow a\)\)and \(\(\overrightarrow b\)\) in component form.

We were supposed to find the vector cross product of the two vectors.

The cross product of two vectors is another vector that is perpendicular to the given vectors. That is, the dot product of the resultant vector and any of the given vectors is zero.

For vectors \(\(\overrightarrow a\)\) and \(\(\overrightarrow b\)\) , the cross product is given as:

\(\[\overrightarrow a \times \overrightarrow b= \begin{vmatrix}\ i & j & k \\a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \end{vmatrix}\]\)

where i, j, k are the unit vectors, a1, a2, a3, b1, b2, b3 are the components of the vectors \(\(\overrightarrow a\) and \(\overrightarrow b\)\)respectively.

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a.There is a 63.21% chance that the light bulb will have to be replaced within 500 hours.

The probability that the light bulb will have to be replaced within 500 hours can be calculated by finding the area under the exponential probability density function (PDF) from 0 to 500. Using the formula for the exponential PDF with a mean of 500, we get:

P(X ≤ 500) = 1 - e^(-500/500) ≈ 0.6321

Therefore, there is a 63.21% chance that the light bulb will have to be replaced within 500 hours.

b. There is a 39.35% chance that the light bulb will last between 200 and 800 hours.

The probability that the light bulb will last more than 1,000 hours can be calculated by finding the area under the exponential PDF from 1000 to infinity. Using the same formula, we get:

P(X > 1000) = e^(-1000/500) ≈ 0.1353

Therefore, there is a 13.53% chance that the light bulb will last more than 1,000 hours.

c. The probability that the light bulb will last between 200 and 800 hours is0.3935.

It can be calculated by finding the area under the exponential PDF from 200 to 800. Again, using the same formula, we get:

P(200 < X < 800) = e^(-200/500) - e^(-800/500) ≈ 0.3935

Therefore, there is a 39.35% chance that the light bulb will last between 200 and 800 hours.

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Answer:

64 square centimeters

Step-by-step explanation:

The surface are of a pyramid is found by finding the sum of the area of the four sides and the base.

Finding the triangular face:

Area of triangle = \(\frac{1}{2} b h\) = \(\frac{1}{2}*4*6 = 12\)

12 * 4 (4 sides) = 48 square cm

Finding the Base = \(w * l = 4 * 4 = 16\)

Finally, we add it together. 48 + 16 = 64

Answer:

b on edge 2021

Step-by-step explanation:

From the calculations, the length of CY in the parallelogram is 15 feet.

A diagonal is a line that runs from one end to another in a quadrilateral. We know that in such quadrilaterals as the parallelogram, the diagonals bisect each other.

Hence;

CY =WC

And

2x - 7 = x + 4

collecting like terms;

2x - x =  4 + 7

x = 11

Thus

CY = 2x - 7

= 2*11 - 7

= 22- 7

= 15

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Answer:

It just asks for the distance so u just have to multiply the two. Standard notation is how it would look as a integer. The answer is 653. 02

Planets around other stars can be detected by carefully measuring the "brightness" or "light intensity" of stars.

When a planet orbits a star, it causes a slight change in the brightness or light intensity of the star. This is known as the transit method of planet detection. As the planet passes in front of the star from our line of sight, it blocks a small portion of the star's light, causing a temporary decrease in its brightness. By carefully measuring these changes in brightness over time, scientists can infer the presence and characteristics of planets orbiting the star. This method has been instrumental in the discovery of numerous exoplanets in recent years.

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Answer:

0.03215.

Step-by-step explanation:

Answer:

Circumference = (3.14)(15) = 47.1 in

Step-by-step explanation:

We get the expression for the statement "The sum of eight and half of a number" as (16 + x) / 2.

We are given a statement:

The sum of eight and half of a number.

We have to form an expression for the same.

An expression means a combination of numbers and variables.

Let the number be a variable say x.

Now we need the half of a number.

1 / 2 of x = x / 2

Now we need the sum of 8 and x / 2.

So, it will be:

8 + x / 2

= (16 + x) / 2

Therefore, we get the expression for the statement "The sum of eight and half of a number" as (16 + x) / 2.

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Answer:

X greater then or equal to -6

Step-by-step explanation:

after doing the math the sign changes because its a nagitive number.

The recursive formula for this sequence for n ≥ 1 , and determine the initial value when n = 1 is:

a_n = a_{n-1} + 3, with a_1 = 12.

Notice that each term in the sequence is 3 more than the previous term. This suggests that the sequence is an arithmetic sequence with a first term of 12 and a common difference of 3.

To write a recursive formula for an arithmetic sequence, we can use the following formula:

a_n = a_{n-1} + d

where a_n is the nth term, a_{n-1} is the previous term, and d is the common difference.

In this case, we have:

a_n = a_{n-1} + 3

with a_1 = 12 as the initial value.

Therefore, the recursive formula for this sequence is:

a_n = a_{n-1} + 3, with a_1 = 12.

This formula can be used in a variety of problems involving arithmetic sequences, such as determining the value of a particular term, finding the sum of a certain number of terms, or determining how long it takes for a certain term to exceed a certain value.

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Using the distance formula of coordinate geometry the expression represents the distance between point J and point K where J= (6, -2) and K= (6, -9) is  7 units.

To find the distance between point J and point K, we use the distance formula, which involves the coordinates of the two points in a coordinate plane.

Identify the coordinates of point J and point K. In this case, we are given that J has coordinates (6, -2) and K has coordinates (6, -9).

Substitute the values of the coordinates into the distance formula, which is d = √[(x2 - x1)² + (y2 - y1)²].

For x1 and y1, use the coordinates of point J, which are x1 = 6 and y1 = -2.

For x2 and y2, use the coordinates of point K, which are x2 = 6 and y2 = -9.

Simplify the formula by substituting the values and solving:

d = √[(6 - 6)² + (-9 - (-2))²]

d = √[0² + (-7)²]

d = √[49]

d = 7

Therefore, the distance between point J and point K is 7 units.

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The Solution of the given expression is -3/2

In mathematics, a fraction is represented by a number that identifies a portion of a whole. A fraction is a component or segment taken from a whole, which can be any number, a certain amount, or an object.  

In fractions, the upper portion is called Numerator and the lower portion is called the Denominator.      

Here given that

fraction numerator x minus 1 over denominator 2 end fraction equals fraction numerator x minus 4 over denominator 5 end fraction

The above fraction can be written Mathematically as given below

=> \([\frac{x-1}{2}] = [\frac{x-4}{5} ]\)

The given expression can be solved as given below

=> \([\frac{x-1}{2}] = [\frac{x-4}{5} ]\)  

=>  \(5(x-1) = 2(x-4)\)

=>  \(5x- 5 = 2x - 8\)

=>  \(5x- 3x = - 8 + 5\)  

=>  \(2x = - 3\)

=>  \(x = -\frac{3}{2}\)  

Therefore,

The Solution of the given expression is -3/2

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Answer: y=8 and x=8\(\sqrt{3}\)

Use the normal 30 60 90 triangle to get y as 16/2, or 8 and x as 16*sqrt3, or

Answer:

Step-by-step explanation:

League A                   League B

151.12                            163.25

148                                157

26.83                             24.93

29                                  136

136                                145

167                                178

207                               256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

\(Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564\)

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=\(\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}\)=\(=\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98\)

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

\((82.5-1.5\times 76.56,159.06+1.5\times 76.56)\)

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

\(Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45\)

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=\(\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}\)=\(=\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42\)

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

\(Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5\)

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

\((140.5-1.5\times 30.125,170.625+1.5\times 30.125)\)

(95.3125,215.8125)

24.93 and 256 are outliers  

Maximum = 256

Answer:

-17/5

Step-by-step explanation:

Multiply ten to everything to get

6x-1= x+2-20

Simplify to 5x= -17

Solve to get x = -17/5

Answer:

"6a = 9"
"x × y = 9"

→ Variables can also be factors in the instance that they are multiplied by a number or another variable.

The correct Mathematical phrases of the English phrases would be j+4, 16c, 3(z+1), ef+8, and v/17 respectively and the solution of given linear equations (part B) will be -105, -291, 270, -48, and -30 respectively.

In part A, the translation of mathematical phrases gives rise to certain equations of linear nature. It helps in easy understanding of the relations between variables and numerical values. In part B, the values of x, b and c are already given for the sake of simplicity. Substituting these values in given linear equations gives the following results:

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Given the values of F (x) at specific points, we need to determine whether there is a local maximum, local minimum, or no extremum at each of these points. The points given are I=0, z=5, x=10, and z=20.

To determine the type of extremum at each point, we can analyze the behavior of the derivative, f'(x), around that point. At I=0: Since the value of f'(x) at x=0 is not given, we cannot make any conclusions about the presence of a local extremum at this point. At z=5: If the derivative f'(x) changes sign from positive to negative as x approaches 5 from the left, then there is a local maximum at x=5. If it changes sign from negative to positive as x approaches 5 from the left, then there is a local minimum at x=5. Without further information about the behavior of f'(x) near x=5, we cannot determine the presence of a local extremum.

At x=10: If the derivative f'(x) changes sign from positive to negative as x approaches 10 from the left, then there is a local maximum at x=10. If it changes sign from negative to positive as x approaches 10 from the left, then there is a local minimum at x=10. Similarly, without knowing the behavior of f'(x) near x=10, we cannot determine the presence of a local extremum.

At z=20: Similar to the previous points, we need information about the behavior of f'(x) near x=20 to determine the presence of a local extremum.  In summary, without additional information about the behavior of f'(x) near the given points, we cannot determine whether there are local maximums, local minimums, or no extremums at I=0, z=5, x=10, and z=20.

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The specific antiderivative of \(F(u)= ln |u|+\frac{u^2}{2} + \frac{5}{2}\)

To find the specific antiderivative \($F(u)$\) of \($f(u)=\frac{1}{u}+u$\) such that \($F(1)=3$\).

First, we find the antiderivative of \($f(u)$\):

\(\int\frac{1}{u}+u du=ln|u|+\frac{u^2}{2}+C\)

where \($C$\) is the constant of integration.

Now, we can use the initial condition. \($F(1)=3$\) to solve for \($C$\):

\(F(1)=ln|1|+\frac{1^2}{2}+C=\frac{1}{2}+C=3\)

Solving for \($C$\), we get \(C=\frac{5}{2}$.\)

The specific antiderivative \($F(u)$\)of \($f(u)$\) is:

\(F(u)= ln |u|+\frac{u^2}{2} + \frac{5}{2}\)

To locate the precise antiderivative \($F(u)$\) of  \($f(u)=\frac{1}{u}+u$\) like that   \($F(1)=3$\).

The antiderivative of\($f(u)$\) is first discovered:

where C is the integration constant.

We may now apply the initial condition.  \($F(1)=3$\) to overcome C:

\(F(1)=ln|1|+\frac{1^2}{2}+C=\frac{1}{2}+C=3\)

The specific antiderivative

\(F(u)= ln |u|+\frac{u^2}{2} + \frac{5}{2}\)

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For the first sample {2.038 2.054 2.053 2.055 2.059 2.059 2.009 2.042 2.053 2.047}, the process is still "in order," while for the second sample {2.022 1.997 2.044 2.044 2.032 2.045 2.045 2.047 2.030 2.044}, the process might be "out of order."

To determine whether the process is still "in order" or "out of order," we can compare the current sample of output to the known mean and standard deviation of the process.

For the first sample {2.038 2.054 2.053 2.055 2.059 2.059 2.009 2.042 2.053 2.047}:

Calculate the sample mean by summing up all the values in the sample and dividing by the number of values (n = 10):

Sample mean = (2.038 + 2.054 + 2.053 + 2.055 + 2.059 + 2.059 + 2.009 + 2.042 + 2.053 + 2.047) / 10 = 2.048.

Compare the sample mean to the known process mean (2.050):

The sample mean (2.048) is very close to the process mean (2.050), indicating that the process is still "in order."

Calculate the sample standard deviation using the formula:

Sample standard deviation = sqrt(sum((x - mean)^2) / (n - 1))

Using the formula with the sample values, we find the sample standard deviation to be approximately 0.019 liters.

Compare the sample standard deviation to the known process standard deviation (0.020):

The sample standard deviation (0.019) is very close to the process standard deviation (0.020), further supporting that the process is still "in order."

For the second sample {2.022 1.997 2.044 2.044 2.032 2.045 2.045 2.047 2.030 2.044}:

Calculate the sample mean:

Sample mean = (2.022 + 1.997 + 2.044 + 2.044 + 2.032 + 2.045 + 2.045 + 2.047 + 2.030 + 2.044) / 10 ≈ 2.034

Compare the sample mean to the process mean (2.050):

The sample mean (2.034) is noticeably different from the process mean (2.050), indicating that the process might be "out of order."

Calculate the sample standard deviation:

The sample standard deviation is approximately 0.019 liters.

Compare the sample standard deviation to the process standard deviation (0.020):

The sample standard deviation (0.019) is similar to the process standard deviation (0.020), suggesting that the process is still "in order" in terms of variation.

In summary, for the first sample, the process is still "in order" as both the sample mean and sample standard deviation are close to the known process values.

However, for the second sample, the difference in the sample mean suggests that the process might be "out of order," even though the sample standard deviation remains within an acceptable range.

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The the function to estimate the average age of employee of a company is A(s)=0.285s + 59  , then the average age of employee in 2003 is 65.27 and in 2009 is 66.98

To estimate the average age of an employee in 2003, we need to find the value of A(s) when s = 22  ;

because the number of years between 2003 and 1981 is = 22 years ;

So , A(22) = 0.285×22 + 59 = 65.27  ;

The average age of an employee in 2003 is approximately 65.27.

To estimate the average age of an employee in 2009,

we need to find the value of A(s) when s = 28

because the number of years between 2009 and 1981 is = 28 years ;

So , A(28) = 0.285×28 + 59 = 66.98  ;

The Average age of employee in 2009 is approximately 71.48.

The given question is incomplete , the complete question is

The function​ A(s) given by ​A(s)=0.285s + 59  can be used to estimate the average age of employee of a company in the year 1981 to 2009. Let​ A(s) be the average age of an​ employee, and "s" be the number of year since​ 1981; that​ is, s=0 for 1981 and s=9 for 1990. What is the average age of the employee in 2003 and in​ 2009 ?

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The statement is true.

When the block is in equilibrium, each spring is stretched an additional ∆x. This implies that the forces from the two springs are balanced, and the block is not experiencing any net force in the equilibrium position.

When the block is set into oscillation with amplitude a, it will pass through its equilibrium point during the oscillation. At the equilibrium point, the displacement of the block is zero, and it changes direction. At this point, the block has its maximum speed v, as it is accelerating towards the equilibrium position.

The speed of the block decreases as it moves away from the equilibrium position, reaches zero at the maximum displacement (amplitude), and then starts accelerating towards the equilibrium point again. Therefore, when the block passes through its equilibrium point, it has its maximum speed v.

To know more about equilibrium refer here:

https://brainly.com/question/14281439?#

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