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3 years ago

Solve for x: 7+4x=13+x

Mathematics

2 answers:

love history [14]3 years ago

5 0

$7+4x=13+x\ \ \ |Subtract\ x\\\\
7+3x=13\ \ \ Subtract\ 7\\\\
3x=6\ \ \ |Divide\ by\ 3\\\\
x=2\\\\
Solution\ is\ x=2.$

Mnenie [13.5K]3 years ago

3 0

$7+4x=13+x \\\\4x-x=13-7 \\\\3x=6 \\\\\boxed{x=\frac{6}{3}=2}$

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What number is five more than the product of six and seven write an expression

vodka [1.7K]

Let us say that this number is n

is means equal sign
product means multiplication
more than means addition
so

$n = 5 + 6 \times 7$

8 0

3 years ago

Complete the chart to find the mean, variance, and standard deviation. Remember to use commas and round numbers to the nearest t

vitfil [10]

Answer: The mean of the data is 433.75, variance of the data is 99667.19 and the standard deviation of the data is 315.7011.

Explanation:

The given data is 900, 35, 500 and 300.

The number of observation is 4.

Formula of mean is,

$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$

$\bar{X}=\frac{900+35+50+300}{4} =433.75$

The formula of variance is given below,

$\sigma^2=\frac{1}{n}\sum{(X-\bar{X})^2}$

$\sigma^2=\frac{398668.75}{4}$

$\sigma^2=99667.19$

The variance of the data is 99667.19

$\sigma=\sqrt{99667.19}$

$\sigma=315.7011$

The standard deviation of the data is 315.7011.

The other information or values of given chart is shown in the attached table.

8 0

3 years ago

Read 2 more answers

Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

5 0

3 years ago

Anyone know this??<br> pls help lol

Svetllana [295]

X={6,-10} that is the answer

3 0

3 years ago

A soccer team traveled to a game in 4 vans. Each van held 6 players. Two of the players are goalkeepers. How many of the players

Pani-rosa [81]

Answer: 22

Work:
4 * 6 = 24
24 - 2 = 22

5 0

3 years ago

Read 2 more answers