Answer:
(1 cm)cos3πt
Step-by-step explanation:
Since the piston starts at its maximal height and returns to its maximal height three times evert 2 seconds, it is modelled by a cosine functions, since a cosine function starts at its maximum point. So, its height h = Acos2πft
where A = amplitude of the oscillation and f = frequency of oscillation and t = time of propagation of oscillation.
Now, since the piston rises in such a way that it returns to the maximal height three times every two seconds, its frequency, f = number of oscillations/time taken for oscillation where number of oscillations = 3 and time taken for oscillations = 2 s
So, f = 3/2 s =1.5 /s = 1.5 Hz
Also, since the the piston moves between 3 cm and 5 cm, the distance between its maximum displacement(crest) of 5 cm and minimum displacement(trough) of 3 cm is H = 5 cm - 3 cm = 2 cm. So its amplitude, A = H/2 = 2 cm/2 = 1 cm
h = Acos2πft
= (1 cm)cos2π(1.5Hz)t
= (1 cm)cos3πt
Answer:
a) 98.522
b) 0.881
c) The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time.
Step-by-step explanation:
a.
As the mentioned in the given instruction the co-variance is first computed in excel by using only add/Sum, subtract, multiply, divide functions.
Marks y Time spent x y-ybar x-xbar (y-ybar)(x-xbar)
77 40 5.1 1.3 6.63
63 42 -8.9 3.3 -29.37
79 37 7.1 -1.7 -12.07
86 47 14.1 8.3 117.03
51 25 -20.9 -13.7 286.33
78 44 6.1 5.3 32.33
83 41 11.1 2.3 25.53
90 48 18.1 9.3 168.33
65 35 -6.9 -3.7 25.53
47 28 -24.9 -10.7 266.43
![Covariance=\frac{sum[(y-ybar)(x-xbar)]}{n-1}](https://tex.z-dn.net/?f=Covariance%3D%5Cfrac%7Bsum%5B%28y-ybar%29%28x-xbar%29%5D%7D%7Bn-1%7D)
Co-variance=886.7/(10-1)
Co-variance=886.7/9
Co-variance=98.5222
The co-variance computed using excel function COVARIANCE.S(B1:B11,A1:A11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted co-variance is 98.52222.
b)
The correlation coefficient is computed as
![Correlation coefficient=r=\frac{sum[(y-ybar)(x-xbar)]}{\sqrt{sum[(x-xbar)]^2sum[(y-ybar)]^2} }](https://tex.z-dn.net/?f=Correlation%20coefficient%3Dr%3D%5Cfrac%7Bsum%5B%28y-ybar%29%28x-xbar%29%5D%7D%7B%5Csqrt%7Bsum%5B%28x-xbar%29%5D%5E2sum%5B%28y-ybar%29%5D%5E2%7D%20%7D)
(y-ybar)^2 (x-xbar)^2
26.01 1.69
79.21 10.89
50.41 2.89
198.81 68.89
436.81 187.69
37.21 28.09
123.21 5.29
327.61 86.49
47.61 13.69
620.01 114.49
sum(y-ybar)^2=1946.9
sum(x-xbar)^2=520.1




The correlation coefficient computed using excel function CORREL(A1:A11,B1:B11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted correlation coefficient is 0.881.
c)
The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time. It means that as the study time increases the marks of student also increases and if the study time decreases the marks of student also decreases.
The excel file is attached on which all the related work is done.
Answer:
A. is your answer.
Step-by-step explanation:
If you look at the picture, you can see that S and T are congruent with each other.
LG and TD are not congruent along with L and D.
A. is your answer.
We can get the answer to this equation by solving for the variable (r).
7.4 - 9 - 2r = 18
We want to get r on one side, so we want to push everything else to the other side.
-2r = 18 - 7.4 + 9
-2r = 19.6
2r = -19.6
r = -9.8