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lana [24]
3 years ago
7

Write each of the following expressions without using absolute value. |z−6|−|z−5|, if z<5

Mathematics
2 answers:
lina2011 [118]3 years ago
6 0

Answer: 1

Step-by-step explanation: first you need to pretend that the absolute value bars are parentheses. Then substitute a with any number less that five, for example z=3

Now we can write our new equation: (3-6)-(3-5)

now we have to determine if the final answer inside the parentheses is positive or negative. In the first parentheses 3-6=-3 with is negative. In our second parentheses we have 3-5=-2 which is a also negative.

Knowing that both parentheses are negative results we can set up an equation using z instead of 3:

-(z-6)-(-(z-5)) is our new equation. If we simplify this equation we get 1 for an answer

Andre45 [30]3 years ago
3 0

Answer: 6 - 5

<u>Step-by-step explanation:</u>

|z - 6| - |z - 5|    ; z < 5

Since z < 5, then

|z - 6| will be the absolute value of a negative number. Replace the absolute value with a negative and parentheses:

-(z - 6) =  -z + 6

|z - 5| will be the absolute value of a negative number. Replace the absolute value with a negative and parentheses:  

-(z - 5) = -z + 5

Now subtract them without the absolute value signs:

-z + 6 - (-z + 5)

Distribute the negative sign:

-z + 6 + z - 5

-z + z = 0 which leaves:

6 - 5

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How many nickels have the same value as a half dollar.?
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8 0
3 years ago
Suppose you roll a fair die. Let X be the value of the roll. What is the Moment Generating Function of X?
Alex_Xolod [135]

Answer:

μ₁`= 1/6

μ₂=  5/36

Step-by-step explanation:

The rolling of a fair die is described by the binomial distribution, as  the

  1. the probability of success remains constant for all trials, p.
  2. the successive trials are all independent
  3. the experiment is repeated a fixed number of times
  4. there are two outcomes success, p, and failure ,q.

The moment generating function of the binomial distribution is derived as below

M₀(t) = E (e^tx)

        = ∑ (e^tx) (nCx)pˣ (q^n-x)

        = ∑ (e^tx) (nCx)(pe^t)ˣ (q^n-x)

        = (q+pe^t)^n

the expansion of the binomial is purely algebraic and needs not to be interpreted in terms of probabilities.

We get the moments by differentiating the M₀(t) once, twice with respect to t and putting t= 0

μ₁`= E (x) = [ d/dt (q+pe^t)^n]  t= 0

            = np

μ₂`=  E (x)² =[ d²/dt² (q+pe^t)^n]  t= 0

              = np +n(n-1)p²

μ₂=μ₂`-μ₁` =npq

in similar way the higher moments are obtained.

μ₁`=1(1/6)= 1/6

μ₂= 1(1/6)5/6

   = 5/36

4 0
3 years ago
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