Answer:
So then the minimum sample to ensure the condition given is n= 38
Step-by-step explanation:
Notation
represent the sample mean for the sample
population mean (variable of interest)
represent the population standard deviation
n represent the sample size
ME = 4 the margin of error desired
Solution to the problem
When we create a confidence interval for the mean the margin of error is given by this formula:
(a)
And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is
. And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got
, replacing into formula (b) we got:
So then the minimum sample to ensure the condition given is n= 38
Answer:
(5,3)
Step-by-step explanation:
Look for the coordinates of the point of intersection where the 2 graphs cross. This gives the solution.
In this case, they cross at (5,3) and they intersect at only one location (i.e there is only 1 solution)
I got 157/990. Just don't ask me how. I just used my fancy TI-84 Calculator.
Answer:
1.6 times or 60% more than
Step-by-step explanation:
The question seems to be asking about the growth factor in the given exponential function.
<h3>Exponential function</h3>
A generic exponential function will have the form ...
quantity = (initial value) × (growth factor)^(number of intervals)
Comparing this form to the given formula ...
f(x) = 58 × 1.6^x
we see the "growth factor" is 1.6. This is the multiplier from one interval (year) to the next.
Each year the expected number of birds is <u><em>1.6 times</em></u> the number the year before.
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Additional comment
A <em>growth factor</em> is sometimes expressed in terms of a <em>growth rate</em>, usually a percentage.
growth factor = 1 + growth rate
1.6 = 1 + 0.60 = 1 + 60%
The growth rate of this bird population is 60% per year. Each year, the population is <u>60% more than</u> the year before.