In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$p = 0.1$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}$

$0.04\sqrt{n} = 0.588$

$\sqrt{n} = \frac{0.588}{0.04}$

$\sqrt{n} = 14.7$

$(\sqrt{n})^{2} = (14.7)^{2}$

$n = 216$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

7 0

3 years ago

Please help me for the brainliest answer

Oxana [17]

Answer:

aₙ= -2n²

Step-by-step explanation:

<h2><u>Solution 1:</u></h2>

The sequence:

-2, -8, -18, -32, -50

The difference between the terms:

-6, -10, -14, -18

a₁= -2
a₂= a₁ - 6 = a₁ - 2*3= a₁- 2*(2²-1)
a₃= a₂ - 10 = a₁ - 16= a₁ - 2*8= a₁ - 2*(3²-1)
a₄= a₃- 14= a₁ - 30= a₁ - 2*15= a₁ - 2*(4² -1)
...
aₙ= a₁ -2*(n²-1)= -2 -2n² +2= -2n²

As per above, the nth term is: aₙ= -2n²

<h2><u /></h2><h2><u>Solution 2</u></h2>

The sequence:

-2, -8, -18, -32, -50
-2*1, -2*4, - 2*9, -2*25
-2*1², -2*2², -2*3², -2*4², -2*5², ..., -2*n²
aₙ= -2n²

6 0

2 years ago

At the middle school graduation dance, the DJ played 12 slow dances, which was equal to the quotient of the number of fast dance

belka [17]

quotient means divide

so 12 = x/2

x = 12*2 = 24

x =24 fast songs

3 0

3 years ago