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nikdorinn [45]
2 years ago
5

PLEASE HELP!! DESPERATE

Mathematics
1 answer:
9966 [12]2 years ago
7 0

a) i) Since blood types across individuals are independent, we have

Pr[1 = A and 2 = B] = Pr[1 = A] • Pr[2 = B]

where e.g. Pr[1 = A] means "probability that partner 1 has type A blood". According to the given distribution,

Pr[1 = A and 2 = B] = 0.37 • 0.13 = 0.0481 = 4.81%

a) ii) The event that partner 1 has type A and partner 2 has type B is exclusive from the event that partner 1 has type B and partner 2 has type A, and vice versa. In other words, only one of these events can happen, and not both simultaneously. This means the probability of either event occurring is equal to the sum of the probabilities of these events occurring individually. Then

Pr[(1 = A and 2 = B) or (1 = B and 2 = A)]

= Pr[1 = A and 2 = B] + Pr[1 = B and 2 = A]

Both probabilities are the same, and we found it in the previous part. So

Pr[(1 = A and 2 = B) or (1 = B and 2 = A)]

= 0.0481 + 0.0481 = 0.0962 = 9.62%

a) iii) The event that at least one partner has type O occurs if exactly one partner has type O, or both do. That is, "at least one partner with type O" contains 3 possible events,

• 1 = O and 2 = not O

• 1 = not O and 2 = O

• 1 = O and 2 = O

For the first event, by independence, is

Pr[1 = O and 2 = not O] = Pr[1 = O] • Pr[2 = not O]

and by definition of complementary event,

Pr[1 = O and 2 = not O] = Pr[1 = O] • (1 - Pr[2 = O])

From the given distribution, it follows that

Pr[1 = O and 2 = not O] = 0.44 • (1 - 0.44) = 0.2464 = 24.64%

The second event is the same as the first in terms of probability.

For the third event, we have

Pr[1 = O and 2 = O] = Pr[1 = O] • Pr[2 = O] = 0.44² = 0.1936 = 19.36%

Then the total probability of at least one O-type partner is

24.64% + 24.64% + 19.36% = 68.64%

b) Suppose partner 1 is type-B. Then partner 2 is an acceptable donor if they are either type-B or type-O. So we want to find

Pr[1 = O and (2 = B or 2 = O)]

By independence,

= Pr[1 = O] • Pr[2 = B or 2 = O]

By mutual exclusivity,

= Pr[1 = O] • (Pr[2 = B] + Pr[2 = O])

Then from the given distribution,

= 0.44 • (0.13 + 0.44) = 0.2508 = 25.08%

Now suppose partner 2 is type-B. Then partner 1 must be either type-B or type-O. But the math works out the same way, so that the overall probability is

25.08% + 25.08% = 50.16%

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The length of a rectangular photo is 1 centimeter less than three times its width. If the area of the photo is 102 square centim
noname [10]

Answers:

  • Length = 17
  • Width = 6

======================================================

How to get those answers:

Let w be the unknown width in centimeters. This variable is some positive real number. This means w > 0 which will be useful later.

The length is "1 cm less than 3 times its width" and it tells us the length is defined by the expression 3w-1. Whatever w is, we triple it to get 3w and then subtract 1 to get the final length.

Multiply the length and width to get the area 102

length*width = area

(3w-1)*w = 102

3w^2-w = 102

3w^2-w-102 = 0

We could guess and check our way to factoring this, but that's not very efficient. The quadratic formula is the better option. It may seem a bit messy, but it's a more direct path that doesn't involve guessing.

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-1)\pm\sqrt{(-1)^2-4(3)(-102)}}{2(3)}\\\\w = \frac{1\pm\sqrt{1225}}{6}\\\\w = \frac{1\pm35}{6}\\\\w = \frac{1+35}{6}\ \text{ or } \ w = \frac{1-35}{6}\\\\w = \frac{36}{6}\ \text{ or } \ w = \frac{-34}{6}\\\\w = 6\ \text{ or } \ w \approx -5.667\\\\

We ignore the second solution (w = -5.667 approximately) because we stated earlier that w > 0. In other words, a negative length does not make sense, so that's why we ignore it.

-----------------

If w = 6 cm is the width, then 3w-1 = 3*6-1 = 18-1 = 17 cm is the length.

Note that length*width = 17*6 = 102 which is the proper area we want. This confirms the answers.

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Answer:

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Step-by-step explanation:

Step 1:

12/13 + ( - 1/3 )

Step 2:

36/39 - 13/39

Answer:

23/39

Hope This Helps :)

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