Answer:
Sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.
Step-by-step explanation:
To yield a more accurate estimate of the population mean, margin of error should be minimized.
margin of error (ME) of the mean can be calculated using the formula
ME=
where
- z is the corresponding statistic in the given confidence level(z-score or t-score)
- s is the standard deviation of the sample (or of the population if it is known)
for a given confidence level, and the same standard deviation, as the sample size increases, margin of error decreases.
Thus, random sample of 50 people from population A, has smaller margin of error than the sample of 20 people from population B.
Therefore, sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.
Answer:
Step-by-step explanation:
3mn (m+y)
= 3
n + 3mny
Answer:
there are 11 big-bags and 24 small-bags.
Step-by-step explanation:
there are x big-bags and y small-bags.
so now we can know:
(1) x + y = 35
(2) 12x + 7y = 300
in (1),we can do like this:
both left and right x7
then (x + y)x7 = 35 x 7
then 7x + 7y = 35 x 7
then 7x + 7y = 245
now,
(1)7x + 7y =245
(2)12x + 7y = 300
we can both left and right do this: (2) - (1)
then
(12x + 7y) - (7x + 7y) = 300 -245
then
12x + 7y - 7x - 7y = 300 -245
then
12x - 7x +7y - 7y =300 -245
then
5x =55
then
5x ÷ 5 = 55 ÷ 5
then
x = 11
because
x + y =35;x=11
so
11 + y =35
11+ y -11 = 35 -11
then
y = 24
now we know:there are 11 big-bags and 24 small-bags.
Answer:
Approximately 3 grams left.
Step-by-step explanation:
We will utilize the standard form of an exponential function, given by:

In the case of half-life, our rate <em>r</em> will be 1/2. This is because 1/2 or 50% will be left after <em>t </em>half-lives.
Our initial amount <em>a </em>is 185 grams.
So, by substitution, we have:

Where <em>f(t)</em> denotes the amount of grams left after <em>t</em> half-lives.
We want to find the amount left after 6 half-lives. Therefore, <em>t </em>= 6. Then using our function, we acquire:

Evaluate:

So, after six half-lives, there will be approximately 3 grams left.