Answer:
65%
Step-by-step explanation:
Given that,
Total fuel = 200000 L
Used fuel = 130,000 L
We need to find the percentage of the used fuel. It can be given by :
So, the required percentage is 65%.
Answer:
(-1.92, 1.08)
Step-by-step explanation:
The <em>incenter</em> is the center of the largest circle that can be inscribed in the triangle. That circle is called the <em>incircle</em>. The incenter is at the point of intersection of the angle bisectors. For a right triangle, the <em>inradius</em> (the radius of the incircle), is found from a simple formula:
r = (a + b - c)/2 . . . . . where c is the hypotenuse, and a and b are the legs
In your triangle, the inradius is ...
r = (5 + 3 -√(5² +3²))/2 = 4 -√8.5 ≈ 1.08452
Among other things, this means the coordinates of the incenter are about (1.08, 1.08) from the right angle vertex, so are about ...
incenter ≈ (-1.92, 1.08)
The third one is 510 not show in the pic here
The correct answer is: " √x − <span>2√b " .
</span>_________________________________________________________
The "conjugate" of " √x + 2√b " is: " √x − 2√b " .
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Explanation:
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In an expression with 2 (TWO) terms; that is, in a "binomial expression",
the "conjugate" of that expression refers to that very expression — with the "sign" in between those two terms—"reverse" (e.g. "minus" becomes "plus" ; or, "plus" becomes "minus" .) .
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→ So: We are given: " <span>√x + 2√b " .
</span>
→ Note that this is a "binomial expression" ;
→ that is, there are 2 (TWO) terms: " <span>√x " ; and: " 2√b " .
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To find the "conjugate" of the given binomial expression:
</span>→ " <span>√x + 2√b " ;
</span>→ We simply change the "+" {plus sign} to a "<span>−" {minus sign} ; and rewrite:
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</span>→ " √x − 2√b " ;
→ which is the "conjugate" ; and is the correct answer:
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→ " √x − 2√b " ; is the "conjugate" of the expression: " <span>√x + 2√b " .
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</span>→ {that is: " √x − 2√b " ; is the conjugate.}.
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The first way to try to fix this is to apply logarithm to the observations on the dependent variable. This is going to make the dependent variable with high degree of kurtosis normal.
Note that sometimes, the resulting values of the variable will be negative. Do not worry about this, as it is not a problem. It does not affect the regression coefficients, it only affects the regression intercept, which after transformation, will be of no interest.
