We are given that there
will be (1/2) a litre after the first pouring, so considering two successive
pourings (n and (n+1)) with 1/2 litre in each before the nth pouring occurs:
1/2 × (1/n) = 1/(2n)
1/2 - 1/(2n) = (n-1)/2n
1/2 + 1/(2n) = (n+1)/2n
(n-1)/2n and (n+1)/2n in
each urn after the nth pouring
Then now consider the
(n+1)th pouring
(n+1)/2n × 1/(n+1) =
1/(2n)
(n+1)/(2n) - 1/(2n) =
n/(2n) = 1/2
Therefore this means that after
an odd number of pouring, there will be 1/2 a litre in each urn
Since 1997 is an odd
number, then there will be 1/2 a litre of water in each urn.
Answer:
<span>1/2</span>
Answer:
b) B and D and c) A and C
Step-by-step explanation:
The hypotheses would be:
(Right tailed test at 5% level)
Sample proportions are
Sample A B C D
Success 31 34 27 38
Proportion p 0.775 0.85 0.675 0.95
Std error
(sqrtpq/n) 0.066 0.056 0.074 0.034
p diff
p-0.75 0.025 0.10 - 0.075 0.20
Z stat
p diff/SE 0.0757 1.77 1.01 5.80
p value 0.469 0.038 0.156 0.000001
we find that p value is more smaller than alpha, the more accurate the alternate hypothesis.
b) Only B and D provide against the null. Because p <0.05
c) A and C provide no evidence for the alternative because p >0.05
y = -2x + 3
y = -x + 10
y = 4x + 10
y = 3/2x - 2
y = -1/2x - 1
y = x + 5
y = 25x + 5
y = .99x + 1.99
y = x - 2
y = -1.50 + 1.50
y = 42x + 27
y = -1/3x - 2
now i am in my phone can see the photo
sometimes in the computer don't appear the picture
Answer:
I got D.
Step-by-step explanation:
Answer:
The exponential parent function with base two is
f(x)=2^x
where x∈(-inf,inf).
when x-> -inf, f(x)-> 0
similarly, when x-> inf, f(x)-> inf
so the range of f(x) is (0,+inf).
Step-by-step explanation:
