Answer:
Step-by-step explanation:
In the diagram below we have
ABCD is a parallelogram. K is the point on diagonal BD, such that
And AK meets BC at E
now in Δ AKD and Δ BKE
∠AKD =∠BKE ( vertically opposite angles are equal)
since BC ║ AD and BD is transversal
∠ADK = ∠KBE ( alternate interior angles are equal )
By angle angle (AA) similarity theorem
Δ ADK and Δ EBK are similar
so we have
( ABCD is parallelogram so AD=BC)
( BC= BE+EC)
( subtracting 1 from both side )
taking reciprocal both side
Answer:
As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.
Step-by-step explanation:
Here n= 20
Sample mean GPA = x`= 2.84
Standard mean GPA = u= 2.55
Standard deviation = s= 0.45.
Level of Significance.= ∝ = 0.01
The hypothesis are formulated as
H0: u1=u2 i.e the GPA of night students is same as the mean GPA of day students
against the claim
Ha: u1≠u2
i.e the GPA of night students is different from the mea GPA of day students
For two tailed test the critical value is z ≥ z∝/2= ± 2.58
The test statistic
Z= x`-u/s/√n
z= 2.84-2.55/0.45/√20
z= 0.1441
As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.
-6x+10 would be ur answer :)
The number solution set 2 . 30 36
Answer:
Less close to one
Step-by-step explanation:
0.006/6,000= 0,000001
