<span>Assume that,
x^2+4x-5 = 0 .......(1)
Then, x^2+4x-5 = 0 x^2+5x-1x-5 =0
x(x+5)-1(x+5) = 0
(x+5) (x-1) = 0
We get x=-5 and x=1
Sub x=-5 in equ (1)
(-5)^2+4(5)-5 = 0
-25+20-5 = 0
-25+25= 0
0 = 0
Sub x=1 in equ (1) (1)^2+4(1)-5 = 0
1+4-5 = 0 5-5 = 0
0 = 0
Therefore x value is -5 and 1</span>

3 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$