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Simora [160]
3 years ago
14

6x-3=3x+12x= What is the value of x in this equation​

Mathematics
2 answers:
goldenfox [79]3 years ago
7 0

Answer:

x = 5

Step-by-step explanation:

6x - 3 = 3x + 12

-3x       -3x

_____________

3x - 3 = 12

    +3    +3

_____________

3x = 15

15 divided by 3 = 5

x = 5

You can double check this by replacing x with (5)

If it is right both sides should equal the same

6(5) - 3 = 3(5) + 12

30 - 3 = 15 + 12

27 = 27

This would prove that x = 5

Leona [35]3 years ago
7 0
The value of x=5 I think
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Please help me on these questions and explain it thanks
BlackZzzverrR [31]
For problem 2:

The answer would be B) Car A travels more miles per gallon of fuel than Car B.

This is because Car B is shown on the graph to travel the same number of miles as Car A using 16 gallons of fuel, while Car A uses only 4 gallons. Thus, Car A travels further with less fuel.

For problem 3:

Let's write out the equation and try to solve.
5x + 1 = 3x + 7

First, subtract 3x from both sides.
5x - 3x + 1 = 3x - 3x + 7
2x + 1 = 7
Now, subtract one from both sides.

2x + 1 - 1 = 7 - 1
2x = 6

Finally, divide both sides by 2.
2x/2 = 6/2

x = 3

You should only get B) One solution

Hope that helped!
8 0
3 years ago
A survey showed that 77% of us need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. 13 adults are randomly
earnstyle [38]

Using the binomial distribution, it is found that the probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.

For each person, there are only two possible outcomes, either they need correction for their eyesight, or they do not. The probability of a person needing correction is independent of any other person, hence, the binomial distribution is used to solve this question.

<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.

In this problem:

  • A survey showed that 77% of us need correction, hence p = 0.77.
  • 13 adults are randomly selected, hence n = 13.

The probability that at least 12 of them need correction for their eyesight is given by:

P(X \geq 12) = P(X = 12) + P(X = 13)

In which:

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

P(X = 12) = C_{13,12}.(0.77)^{12}.(0.23)^{1} = 0.1299

P(X = 13) = C_{13,13}.(0.77)^{13}.(0.23)^{0} = 0.0334

Then:

P(X \geq 12) = P(X = 12) + P(X = 13) = 0.1299 + 0.0334 = 0.163

The probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.

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

7 0
2 years ago
XZ = 13 and YZ = 4 Find XY
prohojiy [21]

Answer:

iddpd

Step-by-step explanation:

kddjdjdkdodidjdjddjdidodododkdjddjdjdjdidodldlddnnxkddo

7 0
3 years ago
The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o
SCORPION-xisa [38]

Answer:

The significance level is \alpha=0.01 and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

z_{\alpha/2}= 2.326

So we reject the null hypothesis is z>2.326

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: \mu \leq 15

Alternative hypothesis: \mu > 15

And the statistic is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

The significance level is \alpha=0.01 and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

z_{\alpha/2}= 2.326

So we reject the null hypothesis is z>2.326

7 0
3 years ago
An equivalent equation has been written by multiplying the second equation by 2.
AVprozaik [17]

its -5 so it is A,( -19/2, -5)

7 0
3 years ago
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