Answer:
The correct answer is:
(a) 0.54
(b) 0.0385
Step-by-step explanation:
Given:
Restaurant tax,
p = 0.54
Sample size,
n = 168
Now,
(a)
The mean will be:
⇒ μ
(b)
The standard error will be:
=
=
=
=
Answer:
Neither, why... for it to be direct or inverse variation, the model has to fit either y=k/x or y=kx, it may not have a y-intercept at all if it is inverse variation and it must have a y-intercept of 0 for it to be direct variation.
Step-by-step explanation:
The statement y=2 is quite specific. Because k is positive, y increases as x increases. So as x increases by 1, y increases by 1.5. Inverse Variation: Because k is positive, y decreases as x increases. yx is a constant number -8. The constant of variation, k , is 23 . Inverse Variation
An inverse variation can be represented by the equation xy=k or y=kx .
That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 .
Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.
91
Follow the order of operations.
3^4 + 2 * 5
81 + 2 * 5 (Exponent)
81 + 10 (Multiplication)
91 (Addition)
Please consider marking this answer as Brainliest to help me advance.
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.