MATHEMATICS COLLEGE

In a sample of 41 water specimens taken from a construction site, 23 contained detectable levels of lead. A 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is 0.409

Required:
Construct a confidence interval for the proportion of water specimens that contain detectable levels of lead.

Answers

Answer 1

Answer:

Answer:

drippy pen iss

Step-by-step explanation:

(8, 1)

for x = -8

$y==-(1)/(8)\cdot(-8)+2=(8)/(8)+2=1+2=3$

(-8, 3)

Mark the points in the coordinate system and draw a straight line through the given points.

(look at the attachment)

An architect uses a scale of 2/3inch to represent 1 foot on a blueprint for a building. If the east wall of the building is 24 feet long, how long (in inches) will the line be on the blueprint?

Answers

Length of the wall on blueprint will be 16 inches.

Use of scale to calculate the distances or length,

Scale used by an architect on a blueprint,

$(2)/(3)\text{ inch}= 1\text{ foot}$

Scale represents the ratio of the length of the wall on blueprint and actual length,

$\frac{\text{Length on the blueprint}}{\text{Actual length}} =\frac{(2)/(3)\text{ inches}}{1\text{ feet}}$

$\frac{\text{Length on the blueprint}}{\text{Actual length}} =(2)/(3)$

If actual length of the east wall of the building = 24 feet

Substitute the value in the expression representing the ratio,

$\frac{\text{Length on the blueprint}}{24} =(2)/(3)$

Length of the blueprint = $(2)/(3)* 24$

= $16$ inches

Therefore, length of the wall on blueprint will be 16 inches.

Learn more about the use of scale to calculate the distances on map.

brainly.com/question/11276271?referrer=searchResults

Answer:

16 in.

Step-by-step explanation:

We have the ratio

$(in.)/(ft): ((2)/(3) )/(1)$

How about let's make this easier. Easier is better, right? Let's get rid of the fraction 2/3. We will do that by multiplying 2/3 by 3 and 1 by 3 to get the equivalent ratio of

$(in.)/(ft):(2)/(3)$

Now we need to know how many inches there would be if the number of feet is 24:

$(in.)/(ft):(2)/(3) =(x)/(24)$

Cross multiply to get

3x = 48 so

x = 16 in.

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

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.

To learn more about the function visit:

brainly.com/question/11624077

#SPJ12

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.