According to the empirical rule, roughly 68% of the scores fall between z = -1 and z = 1
You can use a calculator to get a more approximate answer. If you have a TI calculator, you can use the "normalcdf" function to type in normalcdf(-1,1)
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
Solve for <em>x</em> when √(<em>x</em> ² - 4) = 1 :
√(<em>x</em> ² - 4) = 1
<em>x</em> ² - 4 = 1
<em>x</em> ² = 5
<em>x</em> = ±√5
We're looking at <em>x </em>≤ 0, so we take the negative square root, <em>x</em> = -√5.
This means <em>f</em> (-√5) = 1, or in terms of the inverse of <em>f</em>, we have <em>f</em> ⁻¹(1) = -√5.
Now apply the inverse function theorem:
If <em>f(a)</em> = <em>b</em>, then (<em>f</em> ⁻¹)'(<em>b</em>) = 1 / <em>f '(a)</em>.
We have
<em>f(x)</em> = √(<em>x</em> ² - 4) → <em>f '(x)</em> = <em>x</em> / √(<em>x</em> ² - 4)
So if <em>a</em> = -√5 and <em>b</em> = 1, we get
(<em>f</em> ⁻¹)'(1) = 1 / <em>f '</em> (-√5)
(<em>f</em> ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5
The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of <em>f</em> at <em>x</em> = 1.
Answer:
352
Step-by-step explanation:
:)
:)
:)
The numbers are 1 and 2
1 squared = 1 because 1 x 1 = 1
2 squared = 4 because 2 x 2 = 4
4 - 1 = 3