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Lelechka [254]
3 years ago
14

One number is 9 more than 3 times another. If their difference is 29, what is the larger number?

Mathematics
2 answers:
vfiekz [6]3 years ago
6 0

Answer:

39

Step-by-step explanation:

x-3y=9

x-y=29

subtract

-2y=-20

y=10

x-10=29

x=39

hodyreva [135]3 years ago
3 0

Answer:

<h2>39</h2>

Step-by-step explanation:

x,y-\text{numbers}\\\\\text{One number is 9 more than 3 times another:}\ x=3y+9\\\text{Their difference is equal to 29:}\ x-y=29\\\\\text{We have the system of equations}\\\\\left\{\begin{array}{ccc}x=3y+9&(1)\\x-y=29&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\(3y+9)-y=29\\3y+9-y=29\qquad\text{subtract 9 fro both sides}\\3y-y+9-9=29-9\qquad\text{combine like terms}\\2y=20\qquad\text{divide both sides by 2}\\\dfrac{2y}{2}=\dfrac{20}{2}\\y=10\\\\\text{Put it to (1):}\\x=3(10)+9\\x=30+9\\x=39

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3/8(2x-8/3x)=21 solve the equation for x​
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Explanation:

Multiply Both Sides By 8:

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Simplify:

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Divide Both Sides By -2:

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Step-by-step explanation:

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Determine the average rate of change of the function between the given values of the variable. g(x)=2/x ; x=4, x=a
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3 years ago
if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?
Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

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.

In this problem:

  • There are 9 rolls, hence n = 9.
  • Of the six sides, 2 are 3 or 4, hence p = \frac{2}{6} = 0.3333

The desired probability is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1)

Then

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

P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026

P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0
3 years ago
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