The function A(s) given by A(s)equals0.328splus50 can be used to estimate the average age of employees of a company in the years 1981 to 2009. Let A(s) be the average age of an employee, and s be the number of years since 1981; that is, sequals0 for 1981 and sequals9 for 1990. What was the average age of the employees in 2003 and in 2009?

Answers

Answer 1

Answer:

The average age of the employees in 2003 is 57.216 years. And, the average age of the employees in 2009 is 59.184 years.

Given that;

The function A(s) given by ,

A (s) = 0.328s + 50

Now for the average age of employees in 2003 and 2009 using the function A(s) = 0.328s + 50, substitute the values of s into the equation.

For the year 2003,

Since s represents the number of years since 1981,

Hence, subtract 1981 from 2003:

s = 2003 - 1981

s = 22

Now substitute this value of s into the function A(s):

A(22) = 0.328 × 22 + 50

A(22) = 7.216 + 50

A(22) = 57.216

Therefore, the average age of the employees in 2003 is 57.216 years.

Similarly, for the year 2009,

s = 2009 - 1981

s = 28

Substituting this value into the function:

A(28) = 0.328 × 28 + 50

A(28) = 9.184 + 50

A(28) = 59.184

Hence, the average age of the employees in 2009 is 59.184 years.

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Answer 2

Answer:

Final answer:

The mathematical problem involves calculating the average age of employees at a company for the years 2003 and 2009 using the linear function A(s), where 'A(s)' represents the average age and 's' is the number of years since 1981. The calculated average ages for the employees in the years 2003 and 2009 are approximately 57 and 59 years, respectively.

Explanation:

The subject is mathematics, specifically linear functions. Based on the equation A(s) = 0.328s + 50, where 'A(s)' represents the average age of the employees and 's' represents the number of years since 1981. In the year 2003, s would be 22 (2003-1981) and in 2009, s would be 28 (2009-1981).

Substituting these values of 's' into the function gives:

For 2003, A(22) = 0.328*22 + 50 = 57.216

For 2009, A(28) = 0.328*28 + 50 = 59.184

Therefore, the average age of the employees at the company in 2003 and 2009 were approximately 57 and 59 years, respectively.

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A cable that weighs 6 lb/ft is used to lift 500 lb of coal up a mine shaft 400 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum. (Let x be the distance in feet below the top of the shaft. Enter xi* as xi.)

Answers

The work done to lift the coal is 6.8*10^5 ft-lb

Data;

weight of cable = 6lb/ft
weight of coal = 500lb
length of mine shaft = 400ft

Let the distance (ft) below top of the shaft be represented by x

The weight of the coal to be lifted from mine = 500lb

Work Done

The work done to lift the coal is

$work = force * displacement\nw=fx\nforce f(x) = 500 + 6x\nw_i= f(x_i)=[500+6x_i]\delta x\n$

Taking the summation limit

$w = \lim_(0 \to 400) \sum [500+6x_i]\delta x\n \delta w = f(x) \delta x\n$

we can take the integration of both sides where x will the range from 0 to 400.

$\int \delta w = \int f(x) dx\n\int\limits^(400)_0{(500+6x)} \, dx \nw=[500x+3x^2]_0^4^0^0\nw=500(400-0)+ 3(400^2-0^2)\nw= 680000\nw=6.8*10^5 ft-lb$

From the calculations above, the work done is 6.8*10^5 ft-lb

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Answer:

8.2 *10^5 ft lb

Step-by-step explanation:

Wcoal = 800*500 = 400000 = 4*10^5 ft lb

$Wrope =\int\limits^(400)_0 {6(400-y)} \, dy$

Using a right Riemann sum

The width of the entireregion to be estimated = 400 - 0 = 400

Considering 8 equal subdivisions, then the width of each rectangular division is 400/8 = 50

F(x) = 6(400-y)

$\left[\begin{array}{cccccccccc}x&0&50&100&150&200&250&300&350 &400\nf(x)&2400&2100&1800&1500&1200&900&600&300&0\end{array}\right]$

Wrope = 50(2100) + 50(1800) + 50(1500) + 50(1200) + 50(900) + 50(600)

+ 50(300) + 50(0) = 420000 = 4.2 * 10^5 ft lb

Note: Riemann sum is an approximation, so may not give a accurate value

work done = Wcoal + Wrope= 4*10^5+ 4.2 * 10^5 = 8.2 *10^5 ft lb

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Answers

Answer:

a. As given in the question, some towns are located high in the Rockies above the sea level and some are located in the south-central valley of the California below the sea level as well. The meaningful zero does not exists here and the variable of interest is continuous(quantitative). Hence, the elevations can be best described as an interval data.

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Answers

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Answers

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What are Factors?

Factors of a number or an algebraic expression are the numbers or expressions which divides the given number evenly.

For example, 10 = 2 × 5

So we can say that 2 and 5 are factors of 10.

Greatest common factor of two or more numbers is the factor which is the greatest among all the common factors of both the numbers.

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Answer:

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Answers

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Answers

Answer:

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