No solution
This will help I hope
They already spent $2000, and they only had $5100
Subtract
5100 - 2000 = 3100
$3,050 is your answer, because it is the only one that is less than 3100
hope this helps
Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when 
. So
Rouding up
We need a sample size of at least 719
Answer:
(A) 0.377,
(B) 0.000,
(C) 0.953,
(D) 0.047
Step-by-step explanation:
We assume that having a bone of intention means not liking one's Mother-in-Law
(A) P(all six dislike their Mother-in-Law) = (85%)^6 = (.85)^6 = 0.377
(B) P(none of the six dislike their Mother-in-Law) =
(100% - 85%)^6 =
0.15^6 =
0.000
(C) P(at least 4 dislike their Mother-in-Law) =
P(exactly 4 dislike their Mother-in-Law) + P(exactly 5 dislike their Mother-in-Law) + P(exactly 6 dislike their Mother-in-Law) =
C(6,4) * (.85)^4 * (1-.85)^2 + C(6,5) * (.85)^5 * (.15)^1 + C(6,6) * (.85)^6 = (15) * (.85)^4 * (.15)^2 + (6) * (.85)^5 * .15 + (1) * (.85)^6 =
0.953
(D) P(no more than 3 dislike their Mother-in-Law) =
P(exactly 0 dislikes their Mother-in-Law) + P(exactly 1 dislikes her Mother) + P(exactly 2 dislike their Mother-in-Law) + P(exactly 3 dislike their Mother-in-Law) =
C(6,0) * (.85)^0 * (.15)^6 + C(6,1) * (.85)^1 * (.15)^5 + C(6,2) * (.85)^2 * (.15)^4 + C(6,3) * (.85)^3 * (.15)^3 =
(1)(1)(.15)^6 + (6)(.85)(.15)^5 + (15)(.85)^2 *(.15)^4 + (20)(.85)^3 * (.15)^3 =
0.047
