Question

once again


13 14 15 16 number question plzz solve guys

Answer 1

Q.16 answer is in the img plg chk it

Q. 13

total no. of oranges = 150

probability of its being good =1 - 0.06 =0.94

let the no. of good oranges be x

p [getting good oranges] = x / 150 =0.94

x = 0.94 * 150 =141

no. of good oranges =141

Q. 14

Radius of hemispher (r) = 7cm

Total hieght of Toy = 31cm

Height of Cone (h) = 31 - 7 = 24cm

Slant height of cone (l) =Root of (242 + 72)

Root of (576 + 49)

Root of (625) = 25

Total surface area of toy = CSA of cone + CSA of Hemisphere = 22/7 x r x l + 2 x 22/7 x r2

= 22/7 x r ( l + 2r)

= 22/7 x 7 (25 + 2 x 7)

= 22 x 39 = 858 cm2

Answer 2

13. total no. of oranges = 150

probability of its being good =1 - 0.06 =0.94

let the no. of good oranges be x

p [getting good oranges] = x / 150 =0.94

x = 0.94 * 150 =141

 no. of good oranges =141
14. it is a pic
15. The asked query must be:

a2b2x2- (4b4 - 3a4)x -12a2b2 = 0

⇒ a2 b2x2- 4b4x + 3a4x -12a2b2 = 0

⇒ b2x(a2x - 4b2) + 3a2(a2x - 4b2) = 0

⇒ (a2x - 4b2)(b2x + 3a2) = 0

⇒ (a2x - 4b2) = 0 or (b2x + 3a2) = 0

⇒ a2x = 4b2 or b2x = - 3a2

⇒ x = 4b2/a2 or x = - 3a2/b2
16.
since A.P series is like this

a, (a+d), (a+2d), (a+3d), (a+4d), ……………………………. , {a+(n-1)d}

and nth term is given by = {a+(n-1)d}

so 8th term will be a+7d ,

11th term will be a+10d and

15th term will be a+14d

so as per qn.

a+7d= 31 ……………… (1)

& a+14d = a+10d +16

therefore 4d = 16

& d = 4 ……………….. (2)

use by using (2) in (1)

a + 7 *4 = 31

we get a = 31–28 = 3

so the required A.P will be 3, 7, 11, 15, 19, 23, ……………………
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