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julia-pushkina [17]
3 years ago
11

Y+9÷2=y-3÷4-(y/3)pls help​

Mathematics
2 answers:
faust18 [17]3 years ago
6 0

Step-by-step explanation:

There are 2 scenarios below. One without parenthesis and the second with. I believe the second one is your question.

  • y+9÷2=y-3÷4-(y/3)
  • y - y + y/3 = - 3/4 - 9/2               ⇒ combining like terms
  • y/3 = -3/4 - 18/4
  • y/3 = -21/4
  • y = -21/4*3
  • y = -63/4
  • y = - 15 3/4

===============

  • (y+9)÷2=(y-3)÷4-(y/3)
  • 12*(y+9)/2 = 12* (y-3)/4 - 12*y/3      ⇒ this is to get rid of fraction
  • 6(y+9) = 3(y-3) - 4y
  • 6y + 54 = 3y - 9 - 4y
  • 6y + 54 = - y - 9
  • 6y + y = - 9 - 54
  • 7y = - 63
  • 7y/7 = -63/7
  • y = -9
Ann [662]3 years ago
3 0

Answer:

y = - \frac{63}{4} or - 15.75

Step-by-step explanation:

Order of operations.

y + \frac{9}{2} = y - \frac{3}{4} - \frac{y}{3} Subtract y from each side.

y - y + \frac{9}{2} = y - y - \frac{3}{4} - \frac{y}{3}

\frac{9}{2} =  - \frac{3}{4} - \frac{y}{3}   Add \frac{3}{4} to each side

\frac{9}{2} + \frac{3}{4} = - \frac{3}{4} + \frac{3}{4} - \frac{y}{3}

\frac{9}{2} + \frac{3}{4} = - \frac{y}{3}    Find the common denominator for \frac{9}{2} and \frac{3}{4}, which is 4

\frac{9}{2} * \frac{2}{2} + \frac{3}{4} = - \frac{y}{3}

\frac{18}{4} + \frac{3}{4} = - \frac{y}{3}

\frac{21}{4} = - \frac{y}{3}   Multiply each side by 3

\frac{21}{4}  * 3 = - \frac{y}{3} * 3

\frac{63}{4} = - y      Divide each side by -1

y = - \frac{63}{4} = - 15 \frac{3}{4}

y = - 15.75

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What is the greatest common factor of 45x^3, 24x^2, and 16x?
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Suppose that 37% of college students own cats. If you were to ask random college students if they own a cat what would the proba
Likurg_2 [28]

Using the binomial distribution, the probabilities are given as follows:

a) 0.37 = 37%.

b) 0.5065 = 50.65%.

c) 0.3260 = 32.60%.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37

Item b:

The probability is P(X = 3) when n = 3, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065

Item c:

The probability is P(X = 2) when n = 4, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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2 years ago
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