Answer:
0.3581<x<0.4429
Step-by-step explanation:
Using the formula for calculating the confidence interval of the population proportion p expressed as:
Confidence interval = p ± Z * √p(1-p)/n
p is the population proportion = x/n
p = 200/500
p = 0.4
Z is the z-score at 95% CI = 1.96
n is the sample size = 500
Substituting the given parameters into the formula we will have;
Confidence interval = 0.4 ± 1.96 * √p(1-p)/n
Confidence interval = 0.4 ± 1.96 * √0.4(0.6)/500
Confidence interval = 0.4 ± 1.96 * √0.24/500
Confidence interval = 0.4 ± 1.96 * √0.00048
Confidence interval = 0.4 ± 1.96 * 0.0219
Confidence interval = 0.4±0.04294
Confidence interval = (0.3571, 0.4429)
Hence the confidence interval of the population mean is 0.3581<x<0.4429
Answer:
The answer to this problem is x equals 14
Step-by-step explanation:
A % is out of a hundred, so you don't need the %/100. Then, you wouldn't do cross multiplication, you would just divide 15/27. Remember, it would be 15/27 because it's percent markup, not the percent the price increased by. So, doing that simple calculation on a calculator you get 0.5555. This converted to a percent is just moving the decimal to the right twice, or multiplied by a 100. That would give you 55.55%, and that's your answer. Make sense?
Answer:
1716 ;
700 ;
1715 ;
658 ;
1254 ;
792
Step-by-step explanation:
Given that :
Number of members (n) = 13
a. How many ways can a group of seven be chosen to work on a project?
13C7:
Recall :
nCr = n! ÷ (n-r)! r!
13C7 = 13! ÷ (13 - 7)!7!
= 13! ÷ 6! 7!
(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)
1235520 / 720
= 1716
b. Suppose seven team members are women and six are men.
Men = 6 ; women = 7
(i) How many groups of seven can be chosen that contain four women and three men?
(7C4) * (6C3)
Using calculator :
7C4 = 35
6C3 = 20
(35 * 20) = 700
(ii) How many groups of seven can be chosen that contain at least one man?
13C7 - 7C7
7C7 = only women
13C7 = 1716
7C7 = 1
1716 - 1 = 1715
(iii) How many groups of seven can be chosen that contain at most three women?
(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)
Using calculator :
(15 * 35) + (6 * 21) + (1 * 7)
525 + 126 + 7
= 658
c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?
(First in second out) + (second in first out) + (both out)
13 - 2 = 11
11C6 + 11C6 + 11C7
Using calculator :
462 + 462 + 330
= 1254
d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?
Number of ways with both in the group = 11C5
Number of ways with both out of the group = 11C7
11C5 + 11C7
462 + 330
= 792
They each have $42 because 7 times 6 is 42 and combined they have $84 because 42 + 42=84.
