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:
Vertical Asymptote:

Horizontal asymptote:
it does not exist
Step-by-step explanation:
we are given

Vertical asymptote:
we know that vertical asymptotes are values of x where f(x) becomes +inf or -inf
we know that any log becomes -inf when value inside log is zero
so, we can set value inside log to zero
and then we can solve for x

we get

Horizontal asymptote:
we know that
horizontal asymptote is a value of y when x is +inf or -inf
For finding horizontal asymptote , we find lim x-->inf or -inf



so, it does not exist
Answer:
Part 1)
Bob's mistake was to have used the cosine instead of the sine
The measure of the missing angle is 
Part 2) The surface area of the pyramid is 
Step-by-step explanation:
Part 1)
Let
x----> the missing angle
we know that
In the right triangle o the figure
The sine of angle x is equal to divide the opposite side angle x to the hypotenuse of the right triangle


Bob's mistake was to have used the cosine instead of the sine
Part 2) we know that
The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces
so
![SA=b^{2}+4[\frac{1}{2}(b)(h)]](https://tex.z-dn.net/?f=SA%3Db%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29%5D)
where
b is the length side of the square
h is the height of the triangular lateral face
In this problem
-------> by an 45° angle
so



Find the value of b

Find the surface area
![SA=12^{2}+4[\frac{1}{2}(12)(6)]=288\ cm^{2}](https://tex.z-dn.net/?f=SA%3D12%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%2812%29%286%29%5D%3D288%5C%20cm%5E%7B2%7D)