Complete question :
The computer output and scatter plot output pertaining to the question can be found in the attached picture.
Answer:
Kindly check explanation
Step-by-step explanation:
A.) Relationship between height and fastest serve speed :
According to the scatterplot and Correlation Coefficient (R) value which can be obtauned by getting the square root of R²
R = √27.7%
R = 0.5263 = 52.63% ; this depicts a fairly strong positive correlation or relationship between height and fastest serve speed
B.)
Equation of least square regression line :
From the computer output and scatter plot above :
Recall :
y = mx + c
y =predicted variable (s fastest serve)
m= slope
x = predictor variable (height)
c = intercept
y = 84.98x + 68.81
C.) 27.7% of the variation in fastest speed can be explained by height while the remaining percentage is due to other variables
D.) fastest serve of tennis player with height 1.7m
y = 84.98x + 68.81
y = 84.98(1.7) + 68.81
y = 213.276km/hr
E.)
R = √27.7%
R = 0.5263 = 52.63% ; this depicts a fairly strong positive correlation or relationship between height and fastest serve speed
F.)
Height = 2.06m = x
y = 84.98(2.06) + 68.81
y = 243.8688 km/hr
Residual = Actual - predicted
Actual = 230 ; predicted = 243.8688
Residual = 230 - 243.8688
Residual = −13.8688
Hence, the model overestimated fastest serve speed of a 2.06 m tall player by −13.8688 km/hr
2(5x + 2)^2 = 48
divide both sides by 2
(5x + 2)^2 = 24
square root both sides to clear the power of two.
5x + 2 = + - sqrt24
subtract 2 from both sides
5x = -2 + - sqrt 24
divide both sides by 5
x = (-2 +-sqrt24)/5 or decimal answers .58 and -1.38
<span>Assuming that "pair up students" means "divide up all 20 of the students into groups of two," and we regard two pairings as the same if and only if, in each pairing, each student has the same buddy, then I believe that your answer of 20! / [(2!)^10 * (10!)] is correct. (And I also believe that this is the best interpretation of the problem as you've stated it.)
There are at least two ways to see this (possibly more).
One way is to note that, first, we have to select 2 students for the first pair; that's C(20, 2) (where by C(20, 2) I mean "20 choose 2"; that is, 20! / (18! * 2!). )
Then, for each way of selecting 2 students for the first pair, I have to select 2 of the remaining 18 students for the second pair, so I multiply by C(18, 2).
Continuing in this manner, I get C(20, 2) * C(18, 2) * ... * C(2, 2).
But it doesn't matter in this situation the order in which I pick the pairs of students. Since there are 10! different orders in which I could pick the individual pairs, then I want to divide the above by 10!, giving me the answer
[C(20, 2) * C(18, 2) * ... * C(2, 2)] / 10!.
This is the same as your answer, because C(n , 2) = n(n - 1) / 2, so we can simplify the above as
[(20 * 19) / 2 * (18 * 17) / 2 * ... * (2 * 1) / 2] / 10!
= 20! / [2^10 * 10!]
= 20! / [(2!)^10 * (10!)].
Another way is to reason as follows:
1. First, arrange the 20 students in a line; there are 20! ways to do this
2. We can get a pairing by pairing the 1st and 2nd students in line together, the 3rd and 4th students together, etc.
3. But if I switch the order of the 1st and 2nd student, then this doesn't give a different pairing. I don't want to count both orderings separately, so I divide by 2!
4. The same argument from step 3 holds for the 3rd and 4th student, the 5th and 6th student, etc., so I need to divide by 2! nine more times
5. Finally, the particular order in which I selected the ten pairings are unimportant--for example, the following orderings don't produce different pairings:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
3, 4, 1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
So I need to further divide by the number of ways I can arrange the ten pairs, which is 10!.
</span>
Answer:
a) The function has a maximum at x = 29,59
b) 99,32 %
Step-by-step explanation:
f(x) = -0,11*x² + 6,51*x + 3
Then:
f´(x) = -0,22*x + 6,51
If f´(x) = 0 then - 0,22*x + 6,51 = 0
0,22*x = 6,51
x = 6,51/0,22 ⇒ x = 29,59
if we get the second derivative
f´´(x) = - 0,22 f´´(x) < 0 then we have a maximun at x = 29,59
a) the function has a maximum
b) 29,59 years after 2000 ( in 2030 )
c) f(x) = - 0,11*x² + 6,51*x + 3
f(29,59) = - (0,11)* (29,59)² + 6,51*29,59 + 3
f(29,59) = 99,32 %
Answer:
f
Step-by-step explanation:
A right angled triangle has one angle equal to 90°
an obtuse angle triangle has one angle greater tan 90°
a) An o and right-angled triangle btuse angled triangle can have one angle equal.
b) R and O are not equivalent by definition
c) subset of R and Oare not all triangles as R and O are categories of all triangles. R and O are subset of all triangles
d) Acute triangles have all acute angles. So subset of R and O can't be all acute traingles
e) All triangles with two acute angles are may or may not have third angle as obtuse angle or 90° angle
f) none of the above
