Find the relative percentage error in an error representation of/ by 1.33

Sol:- 

e_p= True value – approx. value × 100

                     True value

True value = 4

                   3

Approx. value= 1.33

So,

True value – approx. value

     = 4 – 1.33

        3

     = 0.0033

So,

  e_p= 0.0033 × 100

         1.3333

      = 0.249975

2. add .4327 E 7 and .0048 E 8

Sol:- 

   =   .4327 E 7 and.0048 E 8

   = .04327 E 8 and .0048 E 8

   = 0.0048 E 8 

3. subtract .8756 E-3 from .4492 E -9  

Sol:- 

       = .8756 E -3 from .0044 E -3

                   = .8712 E -3 

4. Division .6259 E 12 ÷ .0018 E 6

Sol:- 

 = .6259 E^12-6

      .0018 E

   = .6259 E⁶

      .0018

      = 347.72 E 6

         = .34772 E 9

5. Multiplication .9999 E -99 and .1148 E 96

Sol:-

   =  .9999 × .1148 E^-94+96

        = 0.1147 E ⁺²