Question

State which of the following variables are continuous and which are discrete.{chapter:frequency distribution}​

Answer 1

Answer:

for AP:

term1 = a

term3 = a+2d

term9 = a+8d

but term7 = 19 = a+6d

so a = 19-6d

so re-defining:

term1 = 19-6d

term3 = 19-6d+2d = 19-4d

term9 = 19-6d + 8d = 19+2d

these 3 are supposed to be a GP, so

(19-4d)/(19-6d)= (19+2d)/(19-4d)

361 - 152d + 16d^2 = 361 -76d - 12d^2

28d^2 - 76d = 0

d(28d - 76) = 0

d = 0 or d = 76/28 = 19/7

case1 (trivial case) , d = 0

then all terms in the AP would be 19

i.e. 19 19 19 19 ...

of course the first 3 would be a GP also , etc

case 2:

d = 19/7

a = 19 - 6d = 19-6(19/7) = 19/7

for AP, term 20 = a+19d = 19/7+19(19/7) = 380/7

now term1 of AP = term1 of GP

a of GP = 19/7

term 3 of AP = term2 of GP

term2 of GP = a+2d = 19/7 + 2(19/7) = 57/7

so r of GP = (57/7) ÷ (19/7) = 3

sum 12 of GP = a(r^12 - 1)/r

= (19/7)(3^12 - 1)/2

Answer 2

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HERE IS UR ANSWER

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1. discrete
2. discrete
3. continuous
4. discrete
5. discrete
6. continuous
7. discrete
8. continuous
9. discrete