1/8 is the answer jahdbdjdndndndn
3.93 is the correct answer.
I hope this helps.
Answer:
The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

90% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).
On the 1st place he can put 6 textbooks, on the 2nd 5 textbooks etc...
6 · 5 · 4 · 3 = 360
Answer: D) He can arrange textbooks in 360 ways.
Answer:
Step-by-step explanation:
We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.
Our success is winning the prize, p =40/100 = 0.4
Then failure q = 1-0.4 = 0.6
Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.
P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)
=
+
+
+ 
=
+
+
= ![(0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}](https://tex.z-dn.net/?f=%280.6%29%5E%7Bn%7D%2Bn%280.4%29%280.6%29%5E%7Bn-1%7D%2B%5Btex%5D%5Cfrac%7Bn%28n-1%29%280.4%29%5E%7B2%7D0.6%5E%7Bn-2%7D%7D%7B2%7D%20%2B%5Cfrac%7Bn%28n-1%29%28n-2%290.4%5E%7B3%7D0.6%5E%7Bn-3%7D%7D%7B6%7D)