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Liula [17]
3 years ago
12

Let f(x) = 5x3 + 3x2 – 3 and g(x) = x2 + X - 5, fin 4) f(x) + g(x)

Mathematics
1 answer:
aleksley [76]3 years ago
6 0

Answer:

f(x)+g(x)=5x^{3} +4x^{2}+x-8

Step-by-step explanation:

Given:

Two functions are given:

f(x)=5x^{3} +3x^{2} -3

g(x)=x^{2} +x-5

Find f(x)+g(x)

Solution:

f(x)+g(x)=(5x^{3} +3x^{2} -3)+(x^{2} +x-5)

f(x)+g(x)=5x^{3} +3x^{2} -3+x^{2} +x-5

f(x)+g(x)=5x^{3} +(3x^{2} +x^{2})+x-3-5

f(x)+g(x)=5x^{3} +4x^{2}+x-8

Therefore, the function of f(x)+g(x)=5x^{3} +4x^{2}+x-8

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Divide: −72 ÷ 9 = A) −8 B) −7 C) −6 D) 7
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Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.
sergejj [24]

Answer:

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

Step-by-step explanation:

Given that;

fx(x) = { 1/5 ; 50 < x < 100

              0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

                          1 ;   x > 100

a)

n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

Therefore P ( 75 < (x) < 90) =  ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx

= 3/50³ [ -2 (100 - x ]₇₅⁹⁰

= (3 ( -20 + 50)) / 50₃

= 9 / 12500 = 0.00072

b)

f(k) (x) = nf(x)  ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k

Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)

= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

3 0
3 years ago
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