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Llana [10]
3 years ago
7

Doing classes for my ged

Mathematics
2 answers:
shtirl [24]3 years ago
6 0

Answer: 424 oz.

Step-by-step explanation:

1. Multiply the number of large drinks (9) by the amount of ounces in a large drink (32) to get 288.

2. Do the same thing you did with large drinks just use the small drinks numbers (17×8=136)

3. Add the two products (288 and 136) together, which equals 424.

Jlenok [28]3 years ago
4 0

Answer:

424oz

Step-by-step explanation:

you know that 1 large equals 32 oz, so you can make the equation 32*9 which equals 288

you also know that 1 small equals 8 oz, so you can do 8*17 which equals 136

then just add the two answers together. 288+136=424

Hope this helped, feel free to mark brainliest.

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Confidence interval is the range the true values fall in under a given <em>confidence level</em>.

Confidence level states the probability that a random chosen sample performs the surveyed characteristic in the range of confidence interval. Thus,

90% confidence interval means that there is 90% probability that the statistic (in this case SAT score improvement) of a member of the population falls in the confidence interval.

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Mariana has 5 tangerines and decides to give it to the students in her classroom, but since there are 29 students in her classro
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Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
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Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

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$204 in interest

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