Answer:
6.5 hours
Step-by-step explanation:
hope this helped u bye
52x38=1976, 1976 is the answer
37.9 equals 30+7+.9 you should try to put in number form so you know what you're dealing with
Answer:
![6 \sqrt[3]{5}](https://tex.z-dn.net/?f=6%20%5Csqrt%5B3%5D%7B5%7D)
Step-by-step explanation:
For the problem,
, use rules for simplifying cube roots. Under the operations of multiplication and division, if the roots have the same index (here it is 3) you can combine them.
![\sqrt[3]{24} *\sqrt[3]{45} = \sqrt[3]{24*45}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24%7D%20%2A%5Csqrt%5B3%5D%7B45%7D%20%3D%20%5Csqrt%5B3%5D%7B24%2A45%7D)
You can multiply it out completely, however to simplify after you'll need to pull out perfect cubes. Factor 24 and 45 into any perfect cube factors which multiply to each number. If none are there, then prime factors will do. You can group factors together such as 3*3*3 which is 27 and a perfect cube.
![\sqrt[3]{24*45} =\sqrt[3]{3*8*5*3*3} = 6 \sqrt[3]{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24%2A45%7D%20%3D%5Csqrt%5B3%5D%7B3%2A8%2A5%2A3%2A3%7D%20%20%3D%206%20%5Csqrt%5B3%5D%7B5%7D)
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.