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In-s [12.5K]
3 years ago
5

In one type of state lottery 5 distinct numbers are picked from 1,2,3,...,40 uniformly at random.

Mathematics
1 answer:
belka [17]3 years ago
4 0

Answer: a) 1 / ⁴⁰C₅ b)  0.33

Step-by-step explanation:

a) The sample space consists of all numbers 1-40.

Since any of the number can be taken from the sample space  so each of five 5 distinct numbers we take has equal probability of occurring. So probability of each 5 numbers set we take will be equal to 1 / ⁴⁰C₅

b)

If we pick exactly 3 even number then that means other 2 will be odd.

So, we have sample space of 40 numbers out of which 20 are even and 20 are odd.

Now we have to pick 3 even out of 20 and 2 odd out of 20.

Probability = ²⁰C₃ * ²⁰C₂ / ⁴⁰C₅

Probability= 0.33

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Find the value of c so that (x-5) is a factor of the polynomial p(x)
liubo4ka [24]

I think the question is

Find the value of c so that (x-5) is a factor of the polynomial

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The other factor is going to be some quadratic.  We can say a few things about its coefficients but let's start by saying in general it's

q(x)= ax^2 + bx + k

p(x) = (x-5)q(x)

x^3 + 2x^2 + cx + 10 = (x-5)(ax^2 + bx+k) = ax^3 + (b-5a)x^2 + (k-5b)x - 5k

Equating respective coefficients,

a=1

b-5a = 2

k - 5b = c

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so we get

b = 2 + 5 = 7

k = 10/-5 = -2

c = k - 5b = 2 - 5(7)= -37

Answer: -37

Check:

(x^2 + 7x - 2)(x - 5) = x^3 + 2 x^2 - 37 x + 10\quad\checkmark




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