Answer:
a) 30
b) 88
c) 70
d) 16
e) 160
f) 250
Step-by-step explanation:
So the first step would be to switch the numbers around. For example A would be 15 and 50 instead of the original order of 50 and 15. You would then take the 1st number in the new arranged order (15) and multiply it by 100, then divide in that order. 1500/50 = 30.
Another example, letter B.
Step 1 - Rearrange : 22 and 25
Step 2 - Multiple by 100 = 2200
Step 3 - Divide : 2200/25 = 88
Good Luck :) Hope this helped.
The length of a circumference: l=πD
l₁=π×5=5π
l₂=π×15=15π
Then l₂÷l₁=15π÷5π=3 and the answer is B)
Answer:
provides information about the strength of a relationship
Step-by-step explanation:
A numerical measure of strength in the linear relationship between any two variables is called the Pearson's product moment correlation coefficient.
The co efficient of correlation is a pure number denoted by r , independent of the units in which the variables are measured that can range from+1 to -1 .
The sign of r indicates the direction of the cor relation.
When r= 0 it does not mean that there is no relationship . For example if the observed values lie exactly on a circle , there is a relationship between variables but r = 0 as r only measure linear cor relation.
The 2nd statement given is the correct answer.
It is not related to ordinal or nominal properties and it does show direction.
very positive about these results
To prove this inequality we need to consider three cases. We need to see that the equation is symmetric and that switching the variables x and y does not change the equation.
Case 1: x >= 1, y >= 1
It is obvious that
x^y >= 1, y^x >= 1
x^y + y^x >= 2 > 1
x^y + y^x > 1
Case 2: x >= 1, 0 < y < 1
Considering the following sub-cases:
- x = 1, x^y = 1
- x > 1,
Let x = 1 + n, where n > 0
x^y = (1 + n)^y = f_n(y)
By Taylor Expansion of f_e(y) around y = 0,
x^y = f_n(0) + f_n'(0)/1!*y + f_n''(0)/2!*y^2 + ...
= 1 + ln(1 + n)/1!*y + ln(1 + n)^2/2!*y^2 + ...
Since ln(1 + n) > 0,
x^y > 1
Thus, we can say that x^y >= 1, and since y^x > 0.
x^y + y^x > 1
By symmetry, 0 < x < 1, y >= 1, also yields the same.
Case 3: 0 < x, y < 1
We can prove this case by fixing one variable at a time and by invoking symmetry to prove the relation.
Fixing the variable y, we can set the expression as a function,
f(x) = x^y + y^x
f'(x) = y*x^(y-1) + y^x*ln y
For all x > 0 and y > 0, it is obvious that
f'(x) > 0.
Hence, the function f(x) is increasing and hence the function f(x) would be at its minimum when x -> 0+ (this means close to zero but always greater than zero).
lim x->0+ f(x) = 0^y + y^0 = 0 + 1 = 1
Thus, this tells us that
f(x) > 1.
Fixing variable y, by symmetry also yields the same result: f(x) > 1.
Hence, when x and y are varying, f(x) > 1 must also hold true.
Thus, x^y + y^x > 1.
We have exhausted all the possible cases and shown that the relation holds true for all cases. Therefore,
<span> x^y + y^x > 1
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I have to give credit to my colleague, Mikhael Glen Lataza for the wonderful solution.
I hope it has come to your help.
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