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laila [671]
3 years ago
5

Madison planted her garden with 2 5 marigolds and 3 7 tulips. What fraction of the plants in the garden are either marigolds or

tulips?
Mathematics
2 answers:
vfiekz [6]3 years ago
5 0
To answer this you will combine both fractions together to get a total. The denominators are not the same so you will need to create equivalent fractions for both with a common denominator. The denominator is like the units, and to be able to add them they must be the same.

2/5 + 3/7

2/5 = 14/35 and 3/7 = 15/35

14/35 + 15/35 = 29/35

29/35 of the garden are these 2 flowers.

Margaret [11]3 years ago
4 0

Answer:

\frac{29}{35}

Step-by-step explanation:

We have been given that Madison planted her garden with \frac{2}{5} marigolds and \frac{3}{7} tulips.

To find the number of plants in the garden that are either marigolds or tulips, we will add both numbers as:

\frac{2}{5}+\frac{3}{7}

Make a common denominator:

\frac{2\cdot 7}{5\cdot 7}+\frac{3\cdot 5}{7\cdot 5}

\frac{14}{35}+\frac{15}{35}

Add numerators:

\frac{14+15}{35}

\frac{29}{35}

Therefore, \frac{29}{35} of the plants in the garden are either marigolds or tulips.

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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 ?

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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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y = x

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3 years ago
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