In a city of 26,000 population 5000 read English news paper, 12000 read Nepali La News paper and 1000 read both of these. What percent read neither English nor Ne newspaper. ​

38.46%

Step-by-step explanation:

We are given;

Total population; n(S) = 26000

Number that read English news paper; n(E) = 5000

Number of people that read nepali; n(N) = 12000

Number that read both English and nepali; n(E n N) = 1000

Thus from probability laws, number that read either English or nepali is;

n(E u N) = n(N) + n(E) - n(N n E)

n(E u N) = 12000 + 5000 - 1000 = 16000

Thus;

number that read neither English nor nepali is;

n(E' u N') = n(S) - n(E u N)

n(E' u N') = 26000 - 16000 = 10000

Thus, the percentage is;

10000/26000 × 100% = 38.46%

do i do it on a graph?

step-by-step explanation:

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x