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matrenka [14]
3 years ago
11

Hich pairs of fractions include a fraction equivalent to 12 and a fraction equivalent to 25? Select each correct answer. 510 and

48 13 and 412 510 and 410 48 and 410
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

4/8 and 4/10

Step-by-step explanation:

Required

Select which option is equivalent to 1/2 and 2/5

Option (a): 5/10 and 4/8

Simplify 5/10

5/10 = 1/2

Simplify 4/8

4/8 = 1/2

This is not true

Option (b): 1/3 and 4/12

1/3 is not equivalent to 1/2 or 2/5.

So, this is not true

Option (c): 4/12 and 5/10

Simplify 4/12

4/12 = 1/3

1/3 is not equivalent to 1/2 or 2/5.

So, this is not true

Option (d): 4/8 and 4/10

Simplify 4/8

4/8 = 1/2

Simplify 4/10

4/10 = 2/5

Hence, (d) is true

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154

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How many 1 4 cup servings are in 3 8 cups of pudding? (in simplest fraction form) A) 1 6 B) 2 3 C) 3 2 D) 3 8
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Answer: is c 3/2

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3 years ago
A single die is rolled twice. The set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),
lana [24]

Answer:

(9)\frac{1}{12}  (10) \frac{1}{12}  (11)\frac{5}{12}  (12)\frac{1}{4}  (13)\frac{1}{6} 14)\frac{5}{36} (15)\frac{1}{12}  (16)0

Step-by-step explanation:

The sample Space of the single die rolled twice is presented below:

{(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),}.

n(S)=36

(9)Probability of getting two numbers whose sum is 9.

The possible outcomes are:  (3, 6), (4, 5),  (5, 4)

P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are:  (1, 3),(2, 2),(3, 1),

P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}

11.)Find the probability of getting two numbers whose sum is less than 7.

The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4),  (3, 1), (3, 2), (3, 3),  (4, 1), (4, 2),  (5, 1)

P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}

12.Probability of getting two numbers whose sum is greater than 8

The possible outcomes are:(4, 5), (4, 6),  (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}

(13)Probability of getting two numbers that are the same (doubles).

The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4),  (5, 5), (6, 6)

P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}

14.Probability of getting a sum of 7 given that one of the numbers is odd.

The possible outcomes are: (2, 5),  (3, 4), (4, 3), (5, 2),  (6, 1)

P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}

(15)Probability of getting a sum of eight given that both numbers are even numbers.

The possible outcomes are: (2, 6), (4, 4), (6, 2)

P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}

16.Probability of getting two numbers with a sum of 14.

P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0

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Answer:

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Step-by-step explanation:

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