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Solnce55 [7]
2 years ago
14

What situation could this diagram represent?

Mathematics
2 answers:
Aleonysh [2.5K]2 years ago
8 0

Answer: B- Flipping two coins once


Step-by-step explanation:


Darya [45]2 years ago
6 0

B. Tossing Two Coins Once

I Got it Right i Promise

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Suppose we have a group of 10 light users and 10 heavy users. what is the probability that exactly 3 of the 20 users are hiv pos
dmitriy555 [2]

Answer: The answer is 30%.


Step-by-step explanation:  Given that there is a group of 10 light users and 10 heavy users. We are to find the probability that exactly 3 of the 20 users are HIV positive.

We have the following four possible cases -

(i) All 3 are light users.

(ii) 1 is a light user and 2 are heavy users.

(iii) 2 are light users and 1 is a heavy user.

(iv) All 3 are heavy user.

Since there is a 45% chance of a light user to be HIV positive and 55% chance of a heavy user to be HIV positive, so the required probability is given by

p\\\\=\dfrac{3\times 0.45+1\times 0.45+2\times 0.55+2\times 0.45+1\times 0.55+3\times 0.55}{20}\\\\=\dfrac{6}{20}\\\\=0.3\\\\=30\%.

Thus, the probability is 30.

4 0
2 years ago
Brianna bought an assortment of 16 apples. A medium-sized apple weighs around 5.2 ounces. Which response gives the best estimate
harkovskaia [24]
83.2 is your answer here
7 0
2 years ago
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Jenna's heart rate is 60 beats per minute. If Jenna's heart beat 747,533 times, how much time has elapsed? Give your answer in d
olganol [36]

Answer:

Days- 9

Hours- 208

Minutes- 12,459

Step-by-step explanation:

For minutes,

747,533 ÷ 60 = 12,458.883 ≈ 12,459 minutes

For hours,

60 minutes to pass an hour so, 60 × 60 = 3600 minutes

747,533 ÷ 3600 = 207.648 ≈ 208 hours

For days,

1 day = 24 hours

1 hour = 3600 beats

24 × 3600 = 86400

747,533 ÷ 86400 = 8.652 ≈ 9 days

4 0
2 years ago
Read 2 more answers
Cho biết tỉ lệ máy tính bảng sử dụng hệ điều hành A là 70%, tỉ lệ máy tính bảng sử
k0ka [10]

Answer:

máy ...

xác suất để một máy tính bảng có hệ điều hành B sử dụng ổn định trong 2 năm đầu tiên. Add answer

8 0
2 years ago
Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

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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