<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%5Csqrt%7Bx-1%7D%20%7D%20-%5Csqrt%7Bx-2%5Csqrt%7Bx-1%7D%20%7D%20%3D2" id="TexFo

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3 years ago

rmula1" title="\sqrt{x+2\sqrt{x-1} } -\sqrt{x-2\sqrt{x-1} } =2" alt="\sqrt{x+2\sqrt{x-1} } -\sqrt{x-2\sqrt{x-1} } =2" align="absmiddle" class="latex-formula">

with explanation pleaseeee

Mathematics

2 answers:

Nataly [62]3 years ago

6 0

Good morning ☕️

Answer:

Step-by-step explanation:

our expression is defined when x ≥ 1 ,

because , when x ≥ 1

x - 1 ≥ 0

x + 2√(x - 1) ≥ 0

x - 2√(x - 1) ≥ 0

$Calculations:\\\\(\sqrt{x+2\sqrt{x-1} } -\sqrt{x-2\sqrt{x-1} } )^{2} =2^2\\\\then\\\\(x+2\sqrt{x-1} ) +(x-2\sqrt{x-1} ) -2(\sqrt{x+2\sqrt{x-1} })(\sqrt{x-2\sqrt{x-1} } )=4\\\\then\\\\2x - 2\sqrt{(x+2\sqrt{x-1} )(x-2\sqrt{x-1} ) }=4 \\\\then\\\\2x -2\sqrt{(x-2)^2} = 4\\\\then\\\\2x -2|x - 2| =4\\\\then\\\\x - |x-2|=2\\\\now,\\\\case1 : (x\geq2)\\ x - |x-2|=2 \\means \\x-(x-2)=2\\means\\2=2\\\\case2: (1 \leq x \leq2)\\x - |x-2|=2 \\means \\x-(2-x)=2\\means\\2x-2=2\\means\\x=2$

According to case 1 and 2 the solution set for this equality is {x real number where x ≥ 2 }.

zzz [600]3 years ago

6 0

Answer:

Step-by-step explanation:

hello :

look this solution :

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2/3< x/8 <7/8 what number could replace x to make a true statement<br><br> 3<br> 4<br> 5<br> 6

ICE Princess25 [194]

Answer:

Step-by-step explanation:

Multiply all three sides by 8 to get x by itself

16/3 < x < 7

5 1/3 < x < 7

so it’s between 5 and 1/3 and 7. the only answer between those two would be 6. 3 and 4 is too less. 5 is also too less.

5 0

3 years ago

Read 2 more answers

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.