Yes, it is possible to have a perfect square binomial. The detailed information is given below.
<h3>What is the perfect square binomial to trinomial?</h3>
In trinomial, there are three terms and in binomial, there are two terms.
Let the perfect square of the binomial will be given as
⇒ (a + b)²
There are only two terms a and b and has a square on it. So the expression (a + b)² is a perfect square binomial.
Open the bracket, then we have
(a + b)² = a² + b² + 2ab
There are only three terms a², 2ab and b² and derived from a perfect binomial square. So the expression a² + b² + 2ab is a perfect square trinomial.
More about the perfect square binomial to trinomial link is given below.
brainly.com/question/14396504
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<span>8(R + 6) - 2R,
8R + 48 - 2R,
The result is 6R + 48</span>
Answer:
![\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%7Bp%7D%7D%3D0.50%5C%5C%5C%5C%5Csigma_%7B%5Chat%7Bp%7D%7D%3D0.0158)
Step-by-step explanation:
The probability distribution of sampling distribution
is known as it sampling distribution.
The mean and standard deviation of the proportion is given by :-
![\mu_{\hat{p}}=p\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%7Bp%7D%7D%3Dp%5C%5C%5C%5C%5Csigma_%7B%5Chat%7Bp%7D%7D%3D%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
, where p =population proportion and n= sample size.
Given : According to a survey, 50% of Americans were in 2005 satisfied with their job.
i.e. p = 50%=0.50
Now, for sample size n= 1000 , the mean and standard deviation of the proportion will be :-
![\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{0.50(1-0.50)}{1000}}=\sqrt{0.00025}\\\\=0.0158113883008\approx0.0158](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%7Bp%7D%7D%3D0.50%5C%5C%5C%5C%5Csigma_%7B%5Chat%7Bp%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B0.50%281-0.50%29%7D%7B1000%7D%7D%3D%5Csqrt%7B0.00025%7D%5C%5C%5C%5C%3D0.0158113883008%5Capprox0.0158)
Hence, the mean and standard deviation of the proportion for a sample of 1000:
![\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%7Bp%7D%7D%3D0.50%5C%5C%5C%5C%5Csigma_%7B%5Chat%7Bp%7D%7D%3D0.0158)
Answer:
18.03 cubic feet
Step-by-step explanation:
Hello,
Step 1
find the volume of the sphere when radius= 5 ft
![v(r)= \frac{3}{4} \pi r^3\\v(5)= \frac{3}{4}\pi (5\ ft)^3\\v(5)= \frac{3}{4}\pi *125\ ft^{3} \\v(5)=294.52\ cubic\ feet\\](https://tex.z-dn.net/?f=v%28r%29%3D%20%5Cfrac%7B3%7D%7B4%7D%20%5Cpi%20r%5E3%5C%5Cv%285%29%3D%20%5Cfrac%7B3%7D%7B4%7D%5Cpi%20%285%5C%20ft%29%5E3%5C%5Cv%285%29%3D%20%5Cfrac%7B3%7D%7B4%7D%5Cpi%20%2A125%5C%20ft%5E%7B3%7D%20%5C%5Cv%285%29%3D294.52%5C%20cubic%5C%20feet%5C%5C)
Step 2
find the volume of the sphere when radius= 5.1 ft
![v(r)= \frac{3}{4}\pi r^3\\v(5.1)= \frac{3}{4} \pi (5.1\ ft)^3\\v(5.1)= \frac{3}{4}\pi *132.651\ ft^{3} \\v(5.1)=312.551\ cubic\ feet\\\\](https://tex.z-dn.net/?f=v%28r%29%3D%20%5Cfrac%7B3%7D%7B4%7D%5Cpi%20r%5E3%5C%5Cv%285.1%29%3D%20%5Cfrac%7B3%7D%7B4%7D%20%5Cpi%20%285.1%5C%20ft%29%5E3%5C%5Cv%285.1%29%3D%20%5Cfrac%7B3%7D%7B4%7D%5Cpi%20%2A132.651%5C%20ft%5E%7B3%7D%20%5C%5Cv%285.1%29%3D312.551%5C%20cubic%5C%20feet%5C%5C%5C%5C)
Step 3
Compare the Volumes to find the change
![\frac{v(r_{2})}{v(r_{1})} =\frac{312.551}{294.52} =1.06](https://tex.z-dn.net/?f=%5Cfrac%7Bv%28r_%7B2%7D%29%7D%7Bv%28r_%7B1%7D%29%7D%20%3D%5Cfrac%7B312.551%7D%7B294.52%7D%20%3D1.06)
the volumen of the sphere with radius = 5.1 is 1.06 times bigger than the first one(r=5)
Now, find the change
![change= {v(r_{2})-{v(r_{1})}} \\change=312.551\ cubic\ feet\ -294.52 cubic feet \\ change=18.031\ cubic\ feet](https://tex.z-dn.net/?f=change%3D%20%7Bv%28r_%7B2%7D%29-%7Bv%28r_%7B1%7D%29%7D%7D%20%5C%5Cchange%3D312.551%5C%20cubic%5C%20feet%5C%20-294.52%20cubic%20feet%20%5C%5C%20change%3D18.031%5C%20cubic%5C%20feet)
change=18.03 cubic feet
Have a great day.