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3 years ago

The square of the sum of 7 times x and 3 divided by 4 times the difference of x and 1

Mathematics

2 answers:

nikklg [1K]3 years ago

6 0

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

The square of the sum of 7 times x and 3 divided by 4 times the difference of x and 1.

The square of the sum of 7 times x and 3 would be written as

$(7x+3)^2$

and 4 times the difference of x and 1 is written as

$4(x-1)$

Mathematically, it is expressed as

$\dfrac{(7x+3)^2}{4(x-1)}$

Hence, Third option is correct.

4vir4ik [10]3 years ago

4 0

I believe the 3rd answer down if the information you provided is correct

Squaring the sum on the top, rules out 1 and 2 (not squaring the 7 only)
then 4 times the difference. so I believe 4 (x-1) on the bottom.

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit of 750.a. Calculate the mean a

DENIUS [597]

You can compute both the mean and second moment directly using the density function; in this case, it's

$f_X(x)=\begin{cases}\frac1{750-670}=\frac1{80}&\text{for }670\le x\le750\\0&\text{otherwise}\end{cases}$

Then the mean (first moment) is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710$

and the second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3$

The second moment is useful in finding the variance, which is given by

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3$

You get the standard deviation by taking the square root of the variance, and so

$\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09$

8 0

3 years ago