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3 years ago

Find the value of x in the equation x⁄ 4 + x⁄14 + x⁄17 = 71

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

4 0

Hello from MrBillDoesMath!

Answer:

33769/181

Discussion:

Each term contains "x" so factoring it out gives

x( 1/4 + 1/14 + 1/17) = 71 (*)

Use common factor (17*14*4 = 952) as the denominator to combine terms:

1/4 = (17*14)/ 952 = 238/952

1/14 = (17*4)/952 = 68/952

1/17 = (14*4)/952 = 56/952

so 1/4 + 1/14 + 1/17 = (238 + 68 + 56)/ 952 = 362/952 = 181/476

Substituting in (*) gives

x ( 181/476) = 71 => multiply both sides by 476/181

x = (71 * 476)/181 => 71* 476 =33769

x = 33769/181

Thank you,

MrB

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scoray [572]

Answer:

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Step-by-step explanation:

An easy way to solve this problem is to notice the numerator, 2017^4-2016^4 resembles the special product a^2 - b^2. In this case, 2017^4 is a^2 and 2016^4 is b^2. We can set up equations to solve for a and b:

a^2 = 2017^4

a = 2017^2

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b = 2016^2

Now, the special product a^2 - b^2 factors to (a + b)(a - b), so we can substitute that for the numerator:

We can notice that both the numerator and denominator contain 2017^2 + 2016^2, so we can divide by $\frac{2017^2+2016^2}{2017^2+2016^2}$ which is just one, and will simplify the fraction to just:

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This again is just the special product a^2 - b^2, but in this case a is 2017 and b is 2016. Using this we can factor it:

(2017 + 2016)(2017 - 2016)

And, without using a calculator, this is easy to simplify:

(4033)(1)

4033

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2 years ago

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Answer:

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3 years ago

A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 0.1

Ghella [55]

Answer:

$z=-0.674$

And if we solve for $\mu$ we got

$\mu=12 +0.674*0.4=12.270$

So then the mean is $'mu = 12.270$ for this case.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sqrt{0.16}= 0.4)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>12)=0.85$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.85 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.85

If we use condition (b) from previous we have this:

$P(X$