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Sergeu [11.5K]
2 years ago
6

The Occupational Safety and Health Administration (OSHA) mandates certain regulations that have to be adopted by corporations. P

rior to the implementation of the OSHA program, a company found that for a sample of 40 randomly selected months, the mean employee time lost due to job-related accidents was 45 hours. After implementation of the OSHA program, for a random sample of 45 months, the mean employee time lost due tojob-related accidentswas 39 hours. It can be assumed thatthe variability of time lost due to accidentsis about the same before and after implementation of the OSHA program (with a standard deviation being 3.5 hours). (a) Find a 90% confidence interval for the difference in the mean time lost due to accidents. (b) Test the hypothesis that implementation of the OSHA program has reduced the mean employee lost time. Use a level of significance of 0.10
Mathematics
1 answer:
kipiarov [429]2 years ago
6 0

Answer:

Null HYpothesis : H0 : Mean time lost due to job - related accidents are same as after implementing OSHA program as earlier.  

Alternative Hypothesis :

Ha : Mean time lost due to job - related accidents has been decreased after implementing OSHA program

Step-by-step explanation:

In these situations, you must evaluate the employee’s work duties and environment to decide whether or not one or more events or exposures in the work environment either caused or contributed to the resulting condition or significantly aggravated a preexisting condition.

preexisting injury or illness has been significantly aggravated, for purposes of OSHA injury and illness recordkeeping, when an event or exposure in the work environment results in any of the following: (i) Death, provided that the preexisting injury or illness would likely not have resulted in death but for the occupational event or exposure. (ii) Loss of consciousness, provided that the preexisting injury or illness would likely not have resulted in loss of consciousness but for the occupational event or exposure. (iii) One or more days away from work, or days of restricted work, or days of job transfer that otherwise would not have occurred but for the occupational event or exposure. (iv) Medical treatment in a case where no medical treatment was needed for the injury or illness before the workplace event or exposure, or a change in medical treatment was necessitated by the workplace event or exposure

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A circular platform is to be built in a playground. The center of the structure is required to be equidistant from three support
castortr0y [4]

Answer:

The coordinates for the location of the center of the platform are (0, 1)

Step-by-step explanation:

The equation of the circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

Now,

- The center is equidistant from any point lies on the circumference of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

Thus, let's substitute the coordinates of these point in the equation of the circle to find h , k , r.

The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points A(2,−3), B(4,3), and C(−2,5)

- Substitute the values of x and y the coordinates of these points

Point A (2 , -3)

(2 - h)² + (-3 - k)² = r² - - - (1)

Point B (4 , 3)

(4 - h)² + (3 - k)² = r² - - - - (2)

Point C (-2 , 5)

(-2 - h)² + (5 - k)² = r² - - - - (3)

- To find h , k equate equation (1) and (2) and same for equation (2) and (3) because all of them equal r²

Thus;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)² - - - - - (4)

(4 - h)² + (3 - k)² = (-2 - h)² + (5 - k)² - - - - -(5)

- Simplify (5);

h² - 8h + 16 + k² - 6k + 9 = h² + 4h + 4 + k² - 10k + 25

h² and k² will cancel out to give;

-8h - 6k + 25 = 4h - 10k + 29

Rearranging, we have;

12h - 4k = -4 - - - - (6)

Similarly, for equation 4;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)²

h² - 4h + 4 + k² + 6k + 9 = h² - 8h + 16 + k² - 6k + 9

h², k² and 9 will cancel out to give;

4 - 4h + 6k = 16 - 8h - 6k

Rearranging;

4h + 12k = 12 - - - - (7)

Divide by 4 to give;

h + 3k = 3

Making h the subject;

h = 3 - 3k

Put 3 - 3k for h in eq 6;

12(3 - 3k) - 4k = -4

36 - 36k - 4k = -4

40k = 40

k = 40/40

k = 1

h = 3 - 3(1)

h = 0

The coordinates for the location of the center of the platform are (0, 1)

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Why might working on a commission basis make dealing with finances more difficult
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<span>Dealing with finances becomes more difficult when working on a commission basis because unlike working on a salary basis, there is no regular pay. A commission means that earnings are based on rate of sale or number of completed tasks. Income may be reduced if you do not sell enough, or fail to complete enough tasks.</span>
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pickupchik [31]
5q + 2d = 270
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5q = 270
q = 54; (0 , 54)
2d = 270
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The equation represents the total number of coins in the piggy bank and (0 , 54) shows the number of quarters in the piggy bank when there are no dimes and (0 , 135) shows the number of dimes when there are no quarters.
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A ring of grass with an area of 314 yd squared surrounds a circular flower bed. Find the width x of the ring of grass.
tangare [24]

Answer:

1. A ring of grass with an area of 314 yd squared surrounds a circular flower bed. Find the width x of the ring of grass.

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2. Harriet has 80 m of fencing materials to enclose three sides of a rectangular garden. She will use the side of her garage as a border for the fourth side. Find the width x of the garden if its area is to be 700 m squared.

3. Sid cuts four congruent squares from the corners of a 30-in.-by-50-in. rectangular piece of cardboard so that it can be folded to make a box. Find the side length s of the squares, given that teh area of the bottom of the box is 200 in squared.

Step-by-step explanation:

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2 years ago
What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?
Pavel [41]

\boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Explanation:</h2>

Using the quadratic formula:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ f(x) = x^2 + 5x + 5 \\ \\ \\ So: \\ \\ a=1 \\ \\ b=5 \\ \\ c=5 \\ \\ \\ x=\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ \\ x=\frac{-5 \pm \sqrt{25-20}}{2} \\ \\ x=\frac{-5 \pm \sqrt{5}}{2} \\ \\ \\ Two \ solutions: \\ \\ \boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Learn more:</h2>

Quadratic functions: brainly.com/question/12164750

#LearnWithBrainly

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