Answer:
Yup P is the right one having 62.26%
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
We can set it up as
P = 33/53
Q = 20/48
R = 54/90
S = 44/83
This is because we are calculating the percent of yellow birds in the total amt. of birds in a specified park.
Now we calculate =>
P = 33/53 = around 0.62
Q = 20/48 = around 0.416
R = 54/90 = 0.6
S = 44/83 = around 0.53
We find that Park P has the greatest percentage and -->
Thus, Park P is our answer and yes, you are correct.
For the triangle show, what are the values of x and y (urgent help needed)
we just have to use the Pythagoras theorem and then calculate the value of x and y.
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.05 for the estimation of a population proportion
Answer:
A sample of 385 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
How large a sample:
We need a sample of n.
n is found when M = 0.05.
We dont know the true proportion, so we work with the worst case scenario, which is [tex]\pi = 0.5[/tex]
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.05}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.05})^{2}[/tex]
[tex]n = 384.16[/tex]
Rounding up
A sample of 385 is needed.