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serg [7]
3 years ago
9

Can someone hurry please tell me the answer I will mark brainless

Mathematics
1 answer:
Licemer1 [7]3 years ago
7 0

Answer:

C) z = 1/2

Step-by-step explanation:

4z + 5 = 7

4z = 2

z = 1/2

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Please show the work
Alina [70]
The answers are
-12
4
-2

8 0
4 years ago
Need anyone’s help with this APEX question! Much appreciated!
VMariaS [17]

<u>Answer:</u>

The correct answer option is D. (x - 7) + 2.

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

We are to determine whether which of the given options best describe the English expression:

<em>'two more than the quantity of a number minus seven'</em>

We can break this expression into smaller chunks to make it easier to translate.

1. two more than ---> +2

2. the quantity of a number ---> suppose the number to be 'x'

3. number minus 7 ---> (x - 7)

Combining these, we get (x - 7) + 2 so the correct answer option is D.

7 0
4 years ago
Read 2 more answers
Just want to know what is 21x+2
slavikrds [6]

Answer:

21x+2

Step-by-step explanation:

You don't know what the value of x is you can't solve the expression

8 0
3 years ago
The least common denominator of two fractions is 30. If you add the two denominators, their sum is 17. What are the denominators
Ann [662]
X+y=17
x*y=30
It should be 15 and 2
5 0
3 years ago
The n candidates for a job have been ranked 1, 2, 3,..., n. Let X 5 the rank of a randomly selected candidate, so that X has pmf
jeka57 [31]

Question:

The n candidates for a job have been ranked 1, 2, 3,..., n.  Let x = rank of a randomly selected candidate, so that x has pmf:

p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n}  \atop {0\ \ \ Otherwise}} \right.

(this is called the discrete uniform distribution).

Compute E(X) and V(X) using the shortcut formula.

[Hint: The sum of the first n positive integers is \frac{n(n +1)}{2}, whereas the sum of their squares is \frac{n(n +1)(2n+1)}{6}

Answer:

E(x) = \frac{n+1}{2}

Var(x) = \frac{n^2 -1}{12} or Var(x) = \frac{(n+1)(n-1)}{12}

Step-by-step explanation:

Given

PMF

p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n}  \atop {0\ \ \ Otherwise}} \right.

Required

Determine the E(x) and Var(x)

E(x) is calculated as:

E(x) = \sum \limits^{n}_{i} \ x * p(x)

This gives:

E(x) = \sum \limits^{n}_{x=1} \ x * \frac{1}{n}

E(x) = \sum \limits^{n}_{x=1} \frac{x}{n}

E(x) = \frac{1}{n}\sum \limits^{n}_{x=1} x

From the hint given:

\sum \limits^{n}_{x=1} x =\frac{n(n +1)}{2}

So:

E(x) = \frac{1}{n} * \frac{n(n+1)}{2}

E(x) = \frac{n+1}{2}

Var(x) is calculated as:

Var(x) = E(x^2) - (E(x))^2

Calculating: E(x^2)

E(x^2) = \sum \limits^{n}_{x=1} \ x^2 * \frac{1}{n}

E(x^2) = \frac{1}{n}\sum \limits^{n}_{x=1} \ x^2

Using the hint given:

\sum \limits^{n}_{x=1} \ x^2  = \frac{n(n +1)(2n+1)}{6}

So:

E(x^2) = \frac{1}{n} * \frac{n(n +1)(2n+1)}{6}

E(x^2) = \frac{(n +1)(2n+1)}{6}

So:

Var(x) = E(x^2) - (E(x))^2

Var(x) = \frac{(n+1)(2n+1)}{6} - (\frac{n+1}{2})^2

Var(x) = \frac{(n+1)(2n+1)}{6} - \frac{n^2+2n+1}{4}

Var(x) = \frac{2n^2 +n+2n+1}{6} - \frac{n^2+2n+1}{4}

Var(x) = \frac{2n^2 +3n+1}{6} - \frac{n^2+2n+1}{4}

Take LCM

Var(x) = \frac{4n^2 +6n+2 - 3n^2 - 6n - 3}{12}

Var(x) = \frac{4n^2 - 3n^2+6n- 6n +2  - 3}{12}

Var(x) = \frac{n^2 -1}{12}

Apply difference of two squares

Var(x) = \frac{(n+1)(n-1)}{12}

3 0
3 years ago
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