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Butoxors [25]
3 years ago
13

Is the portion 15 / 3 equals 2y / 7 what number is the y in the proportion

Mathematics
2 answers:
GuDViN [60]3 years ago
4 0

Answer:

y = 105 ÷ 6 = 35/2

Step-by-step explanation:

∵ 15/3 = 2y/7

∴ 2y × 3 = 15 × 7

∴ 6y = 105

∴ y = 105 ÷ 6 = 35/2

UNO [17]3 years ago
3 0
<h2>Answer:</h2>

<u>The value of y is </u><u>17.5</u>

<h2>Step-by-step explanation:</h2>

See the attachment

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Algebra 2 help please
suter [353]

Blank 1: 3

Blank 2: 27

Blank 3: 21

you have to distribute the three to everything inside the parenthesis, and because it's directly in front of the parenthesis it's being multiplied.

3(x - 9) = 3x - 27

Now you need to isolate the y, in this e ample you need to add 6 to both sides

the final equation should be:

y = 3x - 21

7 0
2 years ago
What is the formula of the area of any triangle
MrRissso [65]

Answer:Side x side x side

Step-by-step explanation:

Since it is usually equally sided, it would be SxSxS

4 0
3 years ago
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What is the equation of a line with a slope of ​3​ and a point (3, 1) on the line?
dem82 [27]
The answer is:
y = 3x - 8 

Find b by plugging in the point (x,y) which is (3,1).
1= 3(3)+b
1= 9 + b
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5 0
3 years ago
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The first 2 terms of geometric sequnce are shown 3 and 9. What is the next term in the geometric sequence?
Bad White [126]
The next number is 27.
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8 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
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