Answer:
The null hypothesis is that all the different teaching methods have the same average test scores.
H0: μ1 = μ2 = μ3 = μ4 = μ5
The alternative hypothesis is that at least one of the teaching methods have a different mean.
Ha: at least one mean is different. (μ1 ≠ μi)
Step-by-step explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
For the case above, let μ represent the average test scores for the teaching methods:
The null hypothesis is that all the different teaching methods have the same average test scores.
H0: μ1 = μ2 = μ3 = μ4 = μ5
The alternative hypothesis is that at least one of the teaching methods have a different mean.
Ha: at least one mean is different. (μ1 ≠ μi)
GCF of :2xy and 2xy
2(x+y)
Answer:
=-4/2
Step-by-step explanation:
xp=x1 +k (x2-x1)
=-3+1/3 (2- -3)
=-3 + 1/3 (5/1)
= -3 + 5/3
= -9/3 +5/3
= -4/2
sorry it doesnt look the best :)
<span>Position at t=0.35s is 0.2 m
Velocity at t = 0.35s is -0.2 m/s
Since this is college level mathematics, the use of the word "acceleration" should indicate to you that you've been given the 2nd derivative of a function specifying the location of point a. And since you've been asked for the velocity, you know that you want the 1st derivative of the function. And since you've also been asked for the position, you also want the function itself. So let's calculate the desired anti-derivatives.
f''(t) = -1.08 sin(kt) - 1.44 cos(kt)
The integral of f''(t) with respect to t is:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C
In order to find out what C is, we know that at time t=0, v = 0.36 m/s. So let's plug in the values and see what C is:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C
0.36 = (1.08 cos(3*0) - 1.44 sin(3*0))/3 + C
0.36 = (1.08 cos(0) - 1.44 sin(0))/3 + C
0.36 = (1.08*1 - 1.44*0)/3 + C
0.36 = 0.36 + C
0 = C
So the first derivative will be f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k
Now to get the actual function by integrating again. Giving:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C
And let's determine what C is:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C
0.16 = (1.08 sin(3*0) + 1.44 cos(3*0))/3^2 + C
0.16 = (1.08 sin(0) + 1.44 cos(0))/9 + C
0.16 = (1.08*0 + 1.44*1)/9 + C
0.16 = 1.44/9 + C
0.16 = 0.16 + C
0 = C
So C = 0 and the position function is: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2
So now, we can use out position and velocity functions to get the desired answer:
Position:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2
f(t) = (1.08 sin(3*0.35) + 1.44 cos(3*0.35))/3^2
f(t) = (1.08 sin(1.05) + 1.44 cos(1.05))/9
f(t) = (1.08*0.867423226 + 1.44*0.497571048)/9
f(t) = (0.936817084 + 0.716502309)/9
f(t) = 1.653319393/9
f(t) = 0.183702155
So the position of point a at t=0.35s is 0.2 m
Now for the velocity:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k
f'(t) = (1.08 cos(3*0.35) - 1.44 sin(3*0.35))/3
f'(t) = (1.08 cos(1.05) - 1.44 sin(1.05))/3
f'(t) = (1.08*0.497571048 - 1.44*0.867423226)/3
f'(t) = (0.537376732 - 1.249089445)/3
f'(t) = -0.711712713/3
f'(t) = -0.237237571
So the velocity at t = 0.35s is -0.2 m/s</span>
Answer:
Step-by-step explanation:
The first four terms of geometric series is:
Since, we have given information that the third number is greater than 44 that means:
Above equation can be rewritten as:
Now, using:
Here, a=r,b=1
(1)
The sum of first four terms is:
(2)
Divide equation (2) by (1) we get:
CASE1: When r=2 in
CASE2:When r=3 in
The series becomes:
From CASE1: 
From CASE2: 
