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Hitman42 [59]
2 years ago
5

For which values of k does the system of linear equations have zero, one, or an infinite number of solutions? [Note: not all thr

ee possibilities need occur.] (If the answer is an interval of numbers, enter your answer using interval notation. If an answer does not exist, enter DNE.) 8x1 + x2 = 2 kx1 + 9x2 = 18
Mathematics
1 answer:
anastassius [24]2 years ago
4 0

Answer:

For k = 36, there is 0 solutions

For other values of k, there is 1

We never got infinite solutions on the system.

Step-by-step explanation:

We have 2 unknowns and 2 equations:

E1     8x1 + x2 = 18

E2    k*22 x1 + 9 x2 = 18

If we multiply the E1 by 9 we obtain

E3     72 x1 + 9x2 = 162

if we substract E3 with E2 we obtain

(72- 2k) x1 = 144

Thus,

x1 = 144/ (72-2k)

That is, if 72-2k = 0, otherwise there is no solution. And 72-2k = 0 when k = 72/2 = 36.

If k is not 36, then

x1 = 144/(72-2k) and we can replace this value to obtain x2 by using E1

x2 = 18-8x1 = 18- 8 * (144/72-2k)

Which is a specific number that depends only on k. Thus,

for k = 36, there is 0 solutions

for other values of k, there is unique solution.

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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
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Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

5 0
2 years ago
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