Answer:
a) n = 1037.
b) n = 1026.
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 
The margin of error is:
99% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
(a) Assume that nothing is known about the percentage to be estimated.
We need to find n when M = 0.04.
We dont know the percentage to be estimated, so we use 
, which is when we are going to need the largest sample size.
Rounding up
n = 1037.
(b) Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.
So
Rounding up
n = 1026.
Answer:
- 30 adult tickets,
- 50 kids tickets
Step-by-step explanation:
<u>Given</u>
- Cost of adult ticket = $8.50
- Cost of kids ticket = $7.00
- Number of tickets = 80
- Total cost = $605
Let the adult ticket be a and kids be k
<u>We got equations</u>
- a + k = 80
- 8.5a + 7k = 605
<u>From the first equation we get</u>
<u>Substituting k in the second equation</u>
- 8.5a + 7(80 - a) = 605
- 8.5a - 7a + 560= 605
- 1.5a = 45
- a = 45/1.5
- a = 30
<u>Then finding k</u>
The correct question is
<span>In a circle with a radius of 3 ft, an arc is intercepted by a central angle of 2π/3 radians. What is the length of the arc?
we know that
in a circle
</span>2π radians -----------------> lenght of (2*π*r)
2π/3 radians--------------> X
X=[(2π/3)*(2π*r)]/[2π]=(2π/3)*r
the lenght of the arc=(2π/3)*3=2π ft
the answer is 2π ft
Answer:
least common is 1!
Step-by-step explanation:
plz mark me brainliest
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
