Answer: 0.7752
Step-by-step explanation:
Given : The proportion of all homes purchased in 2004 were considered investment properties estimated by NAR: p = 0.23
Sample size : n= 800
Required sample proportion :
Now , the probability that at least 175 homes are going to be used as investments will be :
[using p-value calculator or z-table]
Hence, the required probability = 0.7752
Question:
If The 3rd term of an A.S 23 is and 5th term is 15 .P What is an A.?
Answer:
A.P is 31, 27, 23 ,19, 15, 11, 7, 3
Step-by-step explanation:
Given:
3rd term = 23
5th term = 15
To Find:
The arithmetic Progression = ?
Solution:
Let
be the first term and d be difference
Then the arithmetic progression will be ,
From the question ,
third term = 13
-------------(1)
Similarly
-----------------(2)
Subtracting (1) from (2) we get
-2d =8
d = -4
Substituting d = 4 in equation 1, we get
= 31 ,(31 +(-4)) ,(31 +2(-4)), (31+3(-4)).........
=> 31, 27, 23 , 19, 15, 11,7,3
Step-by-step explanation:
Q3
(a) If 0 < |x − 1| < δ, then |f(x) − 2| < 2
What this means is, how far can x stray from the x=1 line such that f(x) stays within 2 units of the y=2 line (0 < f(x) < 4).
If we move 2 units left of x=1, we get f(x) = 4.
If we move about 3.5 units right of x=1, we get f(x) = 4.
Therefore, δ can't be more than 2.
(b) If 0 < |x| < δ, then |f(x) − 3| < 1
What this means is, how far can x stray from the x=0 line such that f(x) stays within 1 unit of the y=3 line (2 < f(x) < 4).
If we move 1 unit left of x=0, we get f(x) = 4.
If we move 1 unts right of x=0, we get f(x) = 2.
Therefore, δ can't be more than 1.
Since f(x) isn't continuous within this domain, we can't conclude that the limit exists.
Q4
(a) Yes. If δ = 0.25, then 0.75 < x < 1.25, and f(x) > 200.
(b) No. f(1) = 300, so even if δ = 0, f(x) will be less than 400.
(c) Yes. If δ ≈ 0.1, then 0.9 < x < 1, and f(x) > 450.
Height of the cone having volume as and base area
is 6 cm.
Solution:
Given that volume of cone =
Area of base =
Need to determine height of the cone.
Formula for volume of the cone is as follows
Area of circular base of cone =
Replacing in equation (1), we get
In our case Volume of cone and Area of base
On substituting the values of volume and area in equation 2 we get