Answer:
<em>(IQR) interquartile range: 4.5!</em>
Step-by-step explanation:
8, 10, 11, 13, 14, 14, 14, 16, 17
Median: 14
Lower quartile: 10.5
Upper quartile: 15
Interquartile range: 15 - 10.5 = 4.5
Answer:
125/6(In(x-25)) - 5/6(In(x+5))+C
Step-by-step explanation:
∫x2/x1−20x2−125dx
Should be
∫x²/(x²−20x−125)dx
First of all let's factorize the denominator.
x²−20x−125= x²+5x-25x-125
x²−20x−125= x(x+5) -25(x+5)
x²−20x−125= (x-25)(x+5)
∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx
x²/(x²−20x−125) =x²/((x-25)(x+5))
x²/((x-25)(x+5))= a/(x-25) +b/(x+5)
x²/= a(x+5) + b(x-25)
Let x=25
625 = a30
a= 625/30
a= 125/6
Let x= -5
25 = -30b
b= 25/-30
b= -5/6
x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)
∫x²/(x²−20x−125)dx
=∫125/6(x-25) -∫5/6(x+5) Dx
= 125/6(In(x-25)) - 5/6(In(x+5))+C
Answer:
{x | x<3 or x>1}
Step-by-step explanation:
x+2<5
x<5-2
x< 3
x-7> -6
x> -6+7
x>1
symbol U means union
Answer:
0 cm³
Step-by-step explanation:
the lost of water =
(5×24)/12 x 1/10 × 18000
= 10/10 × 18000 = 18,000 cm³
so, the remaining water in the tub=
18,000 - 18,000 = 0 cm³
(the tub is empty)
Answer:
Train A = 128
Train B = 68
Step-by-step explanation:
We can set up a system of equations for this problem
Let A = # of tons of Train A
Let B = # of tons of Train B
A + B = 196
A = B + 60
Now, we plug in A for the first equation, using substitution
(B+60) + B = 196
2B + 60 = 196
Subtract 60 from both sides
2B = 136
Divide both sides by 2
B = 68
Plug in 68 for B in the 2nd equation
A = 68 + 60
A = 128
Checking work: 128 + 68 = 196 :D hope this helped
