1 year ago

What is the standard error of the sampling distribution of sample proportion?.

Mathematics

1 answer:

storchak [24]1 year ago

5 0

Answer:

√ (p(1-p) / n)

Step-by-step explanation:

Standard Error(SE) of the Sample Proportion: √ (p(1-p) / n). Note: as the sample size increases, the standard error decreases.

Hope this helps you! Feel free to give feedback

You might be interested in

Select the correct answer.

Fudgin [204]

A) an answer that cannot be written as a fraction.

6 0

3 years ago

Plz plz help fast <br> I will mark Brainlyist

Licemer1 [7]

Answer:

forgive me i need points

Step-by-step explanation:

7 0

2 years ago

What are the first three terms of the sequence...plz help pt.1

svetlana [45]

$a_{n} \: = \: 2(a_{n - 1})^{2} \\ a_{1} = 4 \: (given) \\ a_{2} = \: 2(a_{2 - 1})^{2} \\ = 2(a_{ 1})^{2} \\ = 2 \times {4}^{2} \\ = 32 \\ \\ a_{3} = \: 2(a_{3 - 1})^{2} \\ = 2(a_{ 2})^{2} \\ = 2 \times {32}^{2} \\ = 2 \times 1024 \\ = 2048$

so first 3 terms are

4, 32, 2048

8 0

3 years ago

Read 2 more answers

Each big square below represents one whole.<br> What percent is represented by the shaded area?

kirza4 [7]

Answer:

145%. The left one is 100% and then the right is the 45%.

5 0

3 years ago

How many terms are there in a geometric series if the first terms is 2, the common ratio is 4, and the sum of the series is 2,73

arsen [322]

The sum of n terms of a geometric series is given by
$S_{n}=a_{1}\dfrac{r^{n}-1}{r-1}$

Substituting the given numbers, you have
$2730=2\dfrac{4^{n}-1}{4-1}\\\\2730\dfrac{3}{2}=4^{n}-1\\\\4096=4^{n}\\\\\dfrac{\log(4096)}{\log(4)}=n=6$

There are 6 terms in the series.

4 0

3 years ago