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SOLUTION of the Above Given Question..🙌✨

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SOLUTION of the Above Given Question..🙌✨

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PROBABILITY JEE PYQ
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Let the probability of getting head for a biased coin be 1/4. It is tossed repeatedly until a head appears. Let N be the number of tosses required. If the probability that the equation 64x2 + 5Nx + 1 = 0 has no real root is p/q, where p and q are co-prime, then q – p is equal to

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[11 April, 2023 (Shift-II)]

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Ye Dimag me hona chahiye question paper attempt krte time ⏰


Graham's law deals with the relation between:
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  •   Pressure and volume
  •   Density and rate of diffusion
  •   Rate of diffusion and volume
  •   Rate of diffusion and viscosity
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PROBABILITY Short Notes..🙌✨

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FORMULA 👇
Area of TRIANGLE

Volume of a TETRAHEDRON


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Let's Understand a Very Basic Topic :- EXPONENTS ( POWER )

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Ye Videos me Detailed Explanation Hai so, watch it carefully fir koi bhi doubt nhi hoga🤩

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Aap apne Galtiyon se hi to sikhoge 🤌


Answer: A. 2√5 units

Explanation: Let the points be P (2, – 1, 3) and Q (– 2, 1, 3)

By using the distance formula,

PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 2, y1 = – 1, z1 = 3

x2 = – 2, y2 = 1, z2 = 3

PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]

= √[(-4)2 + (2)2 + (0)2]

= √[16 + 4 + 0]

= √20

= 2√5

Therefore, the required distance is 2√5 units.


[ THREE DIMENSIONAL GEOMETRY ] What is the distance between the points (2, –1, 3) and (–2, 1, 3)?
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  •   25 units
  •   4√5 units
  •   √5 units
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Three Dimensional Geometry ||

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