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ankoles [38]
3 years ago
5

On a number line, suppose the coordinate of A is 0, and AR = 9. What are the possible coordinates of the midpoint of AR?

Mathematics
1 answer:
m_a_m_a [10]3 years ago
4 0

Given:

On a number line, suppose the coordinate of A is 0, and AR = 9.

To find:

The possible coordinates of the midpoint of AR.

Solution:

Coordinate of A is 0, and AR = 9.

It means the distance between point A and R is 9 units. It means, point R can be either left side or right side of A. So, location of point R can be either -9 or 9.

If coordinate of R is -9, then midpoint of AR is

\dfrac{0+(-9)}{2}=-4.5

If coordinate of R is 9, then midpoint of AR is

\dfrac{0+(9)}{2}=4.5

Therefore, the possible coordinates of the midpoint of AR are -4.5 and 4.5.

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vazorg [7]

Answer:

a) The relative risk is 1.8\overline{846153}

b) The attributable risk is 0.69 \overline{047619}

c) There is a relationship between exposure and tumor risk

Step-by-step explanation:

The number of exposed rats that develop tumors, a = 42

The number of rats exposed to the carcinogen = 600 rats

The number of exposed rats that did not develop tumors, b = 600 - 42 = 558

The number of not exposed rats in the control group = 350 rats

The number of rats that develop tumors in the control group, c = 13 tumors

The number of not exposed rats that did not develop tumors, d = 350 - 13 = 337  rats

a) The relative risk, RR = a/(a + b)/(c/(c + d))

∴ RR = (42/(42 + 558))/(13/(13 + 337)) = 49/26 = 1.8\overline{846153}

b) The attributable risk = (a - c)/a

∴ The attributable risk = (42 - 13)/42 = 0.69 \overline{047619}

c) The odds ratio = (a·b)/(c·d)

∴ The odds ratio = 42 × 558/(13 × 337) = 23439/4381 ≈ 5.35

Given that the result of attributable risk is positive, there is an indication that there is a higher probability to develop tumor when exposed to the potential carcinogen, therefore, there is a relationship between exposure and tumor risk

7 0
3 years ago
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Answer:

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Two payment options to rent a car: You can pay $20 a day plus 25¢ a mile (Option A) or pay $10 a day plus 50¢ a mile (Option B).
BaLLatris [955]

You must drive more than 40 miles to make option A the cheaper plan

<em><u>Solution:</u></em>

Two payment options to rent a car

Let "x" be the number of miles driven in one day

<em><u>You can pay $20 a day plus 25¢ a mile (Option A)</u></em>

25 cents is equal to 0.25 dollars

OPTION A :  20 + 0.25x

<em><u>You pay $10 a day plus 50¢ a mile (Option B)</u></em>

50 cents equal to 0.50 dollars

Option B: 10 + 0.50x

<em><u>For what amount of daily miles will option A be the cheaper plan ?</u></em>

For option A to be cheaper, Option A must be less than option B

Option A < Option B

20 + 0.25x < 10 + 0.50x

Solve the inequality

Add -0.50x on both sides

20 +0.25x -0.50x < 10 + 0.50x - 0.50x\\\\20 - 0.25x < 10

Add - 20 on both sides,

20 - 0.25x - 20 < 10 - 20\\\\-0.25x < -10

\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}

0.25x > 10

Divide both sides by 0.25

x > 40

Thus you must drive more than 40 miles to make option A the cheaper plan

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3 years ago
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barxatty [35]

On question 9, x = 0.46284330

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