Answer:
The answer is 0.8788
Step-by-step explanation:
<em>From question given, let us recall the following:</em>
<em>We know that Ƶα /2 * √p (1-p)/n</em>
<em>when we use n≤ 5000/10 =500</em>
<em> P = 0.75</em>
<em>The Margin of error = 0.03</em>
<em>Putting this values together we arrive at </em>
<em> Ƶα/2 = 0.03/√0.75 * 0.25/500 </em>
<em>= 1.549</em>
<em>Now,</em>
<em>Ф (1.549) = 0.9394</em>
<em>Therefore the confidence level becomes:</em>
<em> 1- (1-∝)/2 = 0.9394</em>
<em>∝ = 0.8788</em>
<em>The answer is 0.8788</em>
<em />
Answer:
x= -2
Step-by-step explanation:
4x + 2 = 6(-2x - 5)
Distribute
4x+2 = -12x-30
Add 12x to each side
4x+2+12x = -12x-30+12x
16x+2 = -30
Subtract 2 from each side
16x+2-2=-30-2
16x = -32
Divide by 16
16x/16x = -32/16
x = -2
Answer:
The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.
Step-by-step explanation:
Given
For any given function f(z), it is analytic and not constant throughout a domain D
To Prove
The function f(z) is non-constant constant in the neighbourhood lying in D.
Proof
1-Assume that the value of f(z) is analytic and has a constant throughout some neighbourhood in D which is ω₀
2-Now consider another function F₁(z) where
F₁(z)=f(z)-ω₀
3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.
4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.
5-Replacing value of F₁(z) in the above gives:
F₁(z)≡0 in domain D
f(z)-ω₀≡0 in domain D
f(z)≡0+ω₀ in domain D
f(z)≡ω₀ in domain D
So this indicates that the value of f(z) for all values in domain D is a constant ω₀.
This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.
Answer:
0.14 or 0.142
Step-by-step explanation:
Answer:
Umm I'm not entirely sure but I think its the last one
