Answer:
-139
Step-by-step explanation:
Evaluate 1/4 (4 x^3 - 2 y - 2 z^3) y^2 - 16 x^2 where x = 2, y = -5 and z = 3:
(4 x^3 - 2 y - 2 z^3)/4 y^2 - 16 x^2 = (4×2^3 - -5×2 - 2×3^3)/4×(-5)^2 - 16×2^2
(4×2^3 - 2 (-5) - 2×3^3)/4×(-5)^2 = ((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4:
((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4 - 16×2^2
(-5)^2 = 25:
((4×2^3 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2
2^3 = 2×2^2:
((4×2×2^2 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2
2^2 = 4:
((4×2×4 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2
2×4 = 8:
((4×8 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2
3^3 = 3×3^2:
((4×8 - 2 (-5) - 23×3^2) 25)/4 - 16×2^2
3^2 = 9:
((4×8 - 2 (-5) - 2×3×9) 25)/4 - 16×2^2
3×9 = 27:
((4×8 - 2 (-5) - 227) 25)/4 - 16×2^2
4×8 = 32:
((32 - 2 (-5) - 2×27) 25)/4 - 16×2^2
-2 (-5) = 10:
((32 + 10 - 2×27) 25)/4 - 16×2^2
-2×27 = -54:
((32 + 10 + -54) 25)/4 - 16×2^2
| 3 | 2
+ | 1 | 0
| 4 | 2:
(42 - 54 25)/4 - 16×2^2
42 - 54 = -(54 - 42):
(-(54 - 42) 25)/4 - 16×2^2
| 5 | 4
- | 4 | 2
| 1 | 2:
(-12×25)/4 - 16×2^2
(-12)/4 = (4 (-3))/4 = -3:
-3×25 - 16×2^2
2^2 = 4:
-3×25 - 164
-3×25 = -75:
-75 - 16×4
-16×4 = -64:
-64 - 75
-75 - 64 = -(75 + 64):
-(75 + 64)
| 7 | 5
+ | 6 | 4
1 | 3 | 9:
Answer: -139
Answer:
I think the answer might be 0.012
Step-by-step explanation:
Answer:
the corresponding point for function f(x) + 2 would be (12, -8)
Answer:
10 + m ≥ 30
You must spend AT MOST $20 or more
Step-by-step explanation:
Basically you have already spent 10 dollars and you have to pay MORE than or equal to 30. You cannot pay less than because you want the free songs. So the sign has to be ≥ or the inequality will not be true.
10 + m ≥ 30
10 - 10 + m ≥ 30 - 10 ( subtract 10 from each side to isolate m)
m ≥ 20 ( we are left with this )
The answer is :
m ≥ 20
You must spend at most 20 dollars more to get the 3 free cd's. Your welcome :D
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
