Ask question

Ask question

All categories

3 years ago

8

Which value, when placed in the circle, would result in a system of equations with infinitely many solutions?

1 answer:

Arlecino [84]3 years ago

5 0

Answer:

A. -10

Step-by-step explanation:

I think you would do like this:

y=2x-5

what about 2y(2x-5) times 2)

so 2y=4x- (-10)

You might be interested in

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

What is 0.25 of 1000

andrew-mc [135]

2.5 is the answer for this

8 0

3 years ago

Read 2 more answers

Which list shows the temperatures in order from coldest to warmet in fahrenheit?

Zigmanuir [339]

The correct answer to this question would be list C.

5 0

3 years ago

Alyssa can paint the room by herself in 2 hours, Bryce can paint the room in 3 hours, and Chris can paint the room in 4 hours. H

TEA [102]

They can paint the room in 1 hour and 5 minutes.

8 0

3 years ago

1/3(9x+27)-4=1/2(2x+54)

Alexandra [31]

First you need to distribute to everything in the parentheses.
3x+9-4=1x+27
Combine like terms
3x+5=x+27
isolate variable
3x+5=x+27
-x -x
2x+5=27
Whatever you do to one side do to the other so I subtract x from both sides. then you must subtract 5 from both sides

2x+5=27
-5 -5
2x=22
divide by 2
2x=22
/2 /2
x=11
Hope that helped

4 0

4 years ago

Read 2 more answers