I m getting 308.11, there is an additional 0.60 that is added, maybe your source is wrong, or there is a copying error.
The range is the ouutput from inputing the input
basically
25=k²+2k+1 and 64=k²+2k+1
the values that satisfy both equations (not at the same tim) are the valuess that are the domain
solve each
25=k²+2k+1
minus 25 both sides (or recognize the perfect square trinomial, but anyway)
0=k²+2k-24
factor
0=(k+6)(k-4)
set to zero
k+6=0
k=-6
k-4=0
k=4
k=-6 or 4
64=k²+2k+1
minus 64 both sides
0=k²+2k-63
facor
0=(k-7)(k+9)
set to zer
k-7=0
k=7
k+9=0
k=-9
k=-9 or 7
so the domain has the numbers
-9,-6,4,7
it seems we only want the positive square roots so
answer is {4,7} is the domain
Answer:
a. H0:μ1≥μ2
Ha:μ1<μ2
b. t=-3.076
c. Rejection region=[tcalculated<−1.717]
Reject H0
Step-by-step explanation:
a)
As the score for group 1 is lower than group 2,
Null hypothesis: H0:μ1≥μ2
Alternative hypothesis: H1:μ1<μ2
b) t test statistic for equal variances
t=(xbar1-xbar2)-(μ1-μ2)/sqrt[{1/n1+1/n2}*{((n1-1)s1²+(n2-1)s2²)/n1+n2-2}
t=63.3-70.2/sqrt[{1/11+1/13}*{((11-1)3.7²+(13-1)6.6²)/11+13-2}
t=-6.9/sqrt[{0.091+0.077}{136.9+522.72/22}]
t=-3.076
c. α=0.05, df=22
t(0.05,22)=-1.717
The rejection region is t calculated<t critical value
t<-1.717
We can see that the calculated value of t-statistic falls in rejection region and so we reject the null hypothesis at 5% significance level.
$1.52 is the total cost. Good luck and have an amazing day!
let sonia has the (x)
Sandeep would have twice as much as Sonia so (2x)
together they'd have 150.
so, x+2x=150
solve for X
3x=150
X=150/3
So x=50 is what Sonia has,
and Sandeep would have 2(50) =100