Answer:
-6.134 to +6.134
Step-by-step explanation:
given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution
X is Normal with mean =6.1 and std dev = 1 unit
We are to determine the range of values containing 70% of the population of x
We know that normal distribution curve is bell shaped symmetrical about the mean.
So to find 70% range we can use 35% on either side of the mean
Using std normal distribution table the value of z for which probability from 0 to z is 0.35 is 1.034
Hence corresponding x value is

i.e. 70% values lie between
-6.134 to +6.134
so
6s+2l = 28
4s+5l= 37
eliminate 6s+2l=28 by dividing all numbers by 6
u get s= -1/3 l + 14/3
49-1/3l=14/3) + 5l =37
L=5
s= -1/3(5) + 14/3
s=3
l=5
The weekdays
Because there are 5 week days so most likely they will meet up on those days
Given:
The given polynomial is

When it divided by (x+1) gives remainder 0 and when divided by (x-2) gives remainder of 45.
To find:
The values of m and n.
Solution:
According to the remainder theorem, if a polynomial P(x) is divided by (x-c), then the remainder is P(c).
Let the given polynomial be P(x).

When it divided by (x+1) gives remainder 0. So, P(-1)=0.



...(i)
When P(x) is divided by (x-2) gives remainder of 45. So, P(2)=0




...(ii)
Multiply 2 on both sides of (i).
...(iii)
Subtract (iii) from (ii).



Putting m=5.5 in (i), we get



Therefore, the values of m and n are 5.5 and -21.5 respectively.