Sample size, n = 75
Point estimate, p = 52/75 = 0.693
Z at 99.7% confidence interval ≈ 2.96
Population mean interval = p+/- Z*Sqrt [p(1-p)/n]
Substituting;
Population mean interval = 0.693 +/- 2.96*Sqrt [0.693(1-0.693)/75] = 0.693+/-0.158 = (0.535,0.851) or (53.5%,85.1%)
Answer:
(n^3 + 4 n + 4487)/(n + 1)
Step-by-step explanation:
Simplify the following:
(n^3 + 4 n - 2 + 67^2)/(n + 1)
| | 6 | 7
× | | 6 | 7
| 4 | 6 | 9
4 | 0 | 2 | 0
4 | 4 | 8 | 9:
(n^3 + 4 n - 2 + 4489)/(n + 1)
Grouping like terms, n^3 + 4 n - 2 + 4489 = n^3 + 4 n + (4489 - 2):
(n^3 + 4 n + (4489 - 2))/(n + 1)
4489 - 2 = 4487:
Answer: (n^3 + 4 n + 4487)/(n + 1)
Answer:
Second option
Step-by-step explanation:
It is given that, if the area of the rectangle to be drawn is 10 square units
To find the length of the rectangle
From the figure, we get the length of AB = 2 units
Therefore Area = length * breadth = 10
Therefore length = 10/2 = 5 units
To find the points of C and D
we have B(-1, 3)
Point C is 5 units below the point B
Therefore C(-1, 3-5) = C(-1,-2)
we have A(1, 3)
The point D is 5 units below the point B
Therefore D(1, 3-5) = A(1,-2)
Therefore the correct answer is option 2.
180 is the totally amount it can be
Answer:
Let C be the cost price of the article. Then:
C(1+30%)=marked price
So:
C(1+30%)=C(1.3)
=1.3C
If the article is then marked down 35%, then its’ price would be:
1.3C(1–35%)=1.3C(.65)
=0.845 or 84.5% of its’ cost
Then:
0.845–1=-0.155
or, a 15.5% loss on the item
