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brilliants [131]
3 years ago
11

Determine Which ratio forms a proportion with 7/2 by finding a common multiplier.

Mathematics
2 answers:
valkas [14]3 years ago
7 0
B will work very nicely. Just multiply 10/10 by 7/2 and you will get 70/20.
rusak2 [61]3 years ago
7 0

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

The ratio forms a proportion with \frac{7}{2}

By finding a common multiplier

A) \frac{35}{8} can't be written as \frac{7}{2}

C) \frac{28}{10}=\frac{14}{5} can't be written as \frac{7}{2}

D) \frac{70}{8}=\frac{35}{4} can't be written as \frac{7}{2}

But,

B ) \frac{70}{20}=\frac{7}{2} is correct .

Hence, Option 'B' is correct.



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Using a fair coin and a number cube with faces numbered 1 through 6, what is the probability of getting heads and rolling a 5?
torisob [31]

Step-by-step explanation:

Use Multiplication Principle with states that if a event can occur n ways, and another mutually exclusive event occur p ways, then the total outcome is

n \times p

The possible ways of rolling a dice is 6 and the possible ways of flipping a coin is 2 so we have

6 \times 2 = 12

Now, we use the fact that proabliblity is

Number of favorable outcomes/ Total outcomes.

We can only get a head and a 5 once out of the set so the number of favorable outcomes is 1. Total outcomes is 12 so we have

\frac{1}{12}

A is the answer.

3 0
2 years ago
The math club is sponsoring a bake sale.If their goal is to raise at least $200,how many pies must they sell at $4.00 each in or
Tems11 [23]
The math club must sell 50 pies in order to reach the goal of 200 dollars
4 0
3 years ago
g Suppose that a die is rolled twice. What are the possible values that thefollowing random variables can take on:(a) the maximu
KengaRu [80]

Answer:

(a) A = {1, 2, 3, 4, 5, 6}

(b) B = {1, 2, 3, 4, 5, 6}

(c) C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(d) D = {-5, -4, -3, -2, -1, -0, 1, 2, 3, 4, 5}

Step-by-step explanation:

Assume each roll can result in the numbers 1, 2, 3, 4, 5, or 6.

(a) If both rolls result in a 1, the maximum value is 1. If either roll results in a 6, the maximum value is 6; all integers between 1 and 6 are also possible. Therefore, the possible values are:

A = {1, 2, 3, 4, 5, 6}

(b) If either roll results in a 1, the minimum value is 1. If both rolls result in a 6, the minimum value is 6; all integers between 1 and 6 are also possible. Therefore, the possible values are:

B = {1, 2, 3, 4, 5, 6}

(c) If both rolls result in a 1, the sum is 2. If both rolls results in a 6, the sum is 12; all integers between 2 and 12 are also possible. Therefore, the possible values are:

C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(d) If the first roll results in a 1 and the second results in a 6, the result is -5. On the other hand, if the first roll results in a 6 and the second results in a 1, the result is 5; all integers between -5 and 5 are also possible. Therefore, the possible values are:

D = {-5, -4, -3, -2, -1, -0, 1, 2, 3, 4, 5}

7 0
2 years ago
OwOoWoOwO pls answer, sorry about the random burst im really bad at math
Nataly [62]

Answer:

0.46

Step-by-step explanation:

42.78/m = 93

=> m = 42.78/93

=> m = 0.46

Hope my answer helps :)

7 0
2 years ago
Read 2 more answers
A bag contains blue marbles and red marbles, 56 in total. The number of blue marbles is 8 more than 2 times the number of red ma
Nikolay [14]

24 Blue marbles.

Explanation:

56 - 8 = 48

48 / 2 = 24

so if there are 24 blue marbles, 24 x 2 = 48, and then 8 more is 56.

8 0
3 years ago
Read 2 more answers
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