Answer:
![V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%28X%5E2%29-%5BE%28X%29%5D%5E2%3D349.2-%2818.6%29%5E2%3D3.24)
The expected price paid by the next customer to buy a freezer is $466
Step-by-step explanation:
From the information given we know the probability mass function (pmf) of random variable X.
<em>Point a:</em>
- The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is
Therefore
- If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by
![E[h(X)] = $\sum_{D} h(x)\cdot p(x)](https://tex.z-dn.net/?f=E%5Bh%28X%29%5D%20%3D%20%24%5Csum_%7BD%7D%20h%28x%29%5Ccdot%20p%28x%29)
So 
and
![E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2](https://tex.z-dn.net/?f=E%5Bh%28X%29%5D%20%3D%20%24%5Csum_%7BD%7D%20h%28x%29%5Ccdot%20p%28x%29%5C%5CE%5BX%5E2%5D%3D%24%5Csum_%7BD%7Dx%5E2%5Ccdot%20p%28x%29%5C%5C%20E%28X%5E2%29%3D16%5E2%5Ccdot%200.3%2B18%5E2%5Ccdot%200.1%2B20%5E2%5Ccdot%200.6%5C%5CE%28X%5E2%29%3D349.2)
- The variance of X, denoted by V(X), is
![V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2](https://tex.z-dn.net/?f=V%28X%29%20%3D%20%24%5Csum_%7BD%7DE%5B%28X-%5Cmu%29%5E2%5D%3DE%28X%5E2%29-%5BE%28X%29%5D%5E2)
Therefore
![V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%28X%5E2%29-%5BE%28X%29%5D%5E2%5C%5CV%28X%29%3D349.2-%2818.6%29%5E2%5C%5CV%28X%29%3D3.24)
<em>Point b:</em>
We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:
From the rules of expected value this proposition is true:
We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore
The expected price paid by the next customer is
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
<h2>
Answer:</h2>
Option: B is the correct answer.
The range of the function is:
B. 5 < y < ∞
<h2>
Step-by-step explanation:</h2>
Range of a function--
The range of a function is the set of all the values that is attained by the function.
By looking at the graph of the function we see that the function tends to 5 when x→ -∞ and the function tends to infinity when x →∞
Also, the function is a strictly increasing function.
This means that the function takes every real value between 5 and ∞ .
i.e. The range of the function is: (5,∞)
Hence, the answer is:
Option: B
Answer:
of the phones in office has both a speaker phone and a conference-call capability
Step-by-step explanation:
Since total number of phones in the office is not given, the answer will be "in that terms".
let it be x
So built in speaker phones is (1/4)x
Hence conference call phones would be (1/2)(1/4x) = (1/8)x
Hence "1/8th of the phones in office has both a speaker phone and a conference-call capability"
