The fraction is..
1/3
i dont know the decimal tho
Can u put a picture of the question with this so I can answer it.
Answer:
-2(16xy-10y+7)
Step-by-step explanation:
-2(16xy-10y-7)= -2×16xy-2(-10y)-2×7
-2(16xy-10y+7)= -32xy+20y-14
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Group 1:
μ1 = 59.7
s1 = 2.8
n1 = sample size = 12
Group 2:
μ2 = 64.7
s2 = 8.3
n2 = sample size = 15
α = 0.1
Assume normal distribution and equ sample variance
A.)
Null and alternative hypothesis
Null : μ1 = μ2
Alternative : μ1 < μ2
B.)
USing the t test
Test statistic :
t = (m1 - m2) / S(√1/n1 + 1/n2)
S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))
S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))
S = 6.4829005
t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)
t = - 5 / 2.5108165
tstat = −1.991384
Decision rule :
If tstat < - tα, (n1+n2-2) ; reject the Null
tstat < t0.1,25
From t table :
-t0.1, 25 = - 1.3163
tstat = - 1.9913
-1.9913 < - 1.3163 ; Hence reject the Null
Answer:
1. x2 - 9 > 0
x^2-3^2>0
(x+3)(x-3)>0
(x+3)>0 and (x-3)>0
x>-3 and x>3
2. x2 - 8x + 12 > 0
x^2 - 8x +12>0
x^2 -2x -6x +12 >0 (-8x is replaced by (-2x) + (-6x) )
x(x-2) -6(x-2) >0
(x-6)(x-2)>0
(x-6)>0 and (x-2)>0
x>6 and x>2
3. -x2 - 12x - 32 > 0
-x^2 -12x -32 >0
x^2 +12x +32 <0
x^2 +4x +8x +32<0
x(x+4) +8(x+4)<0
(x+8)(x+4)<0
(x+8)<0 and (x+4)<0
x<-8 and x<-4
4. x2 + 3x - 20 >= 3x + 5
x^2 +3x -20 >= 3x +5
x^2 +3x -20 -3x >= 3x +5 -3x
x^2 -20 >= 5
x^2 -20 +20 >= 5 +20
x^2 >=25
x^2-25 >=0
(x-5)(x+5)>=0
(x-5)>=0 and (x+5)>=0
x>=5 and x>=-5
