perimeter means add the side
(x+4) +( x+8)+(2x-3) =P
x+x+2x+4+8-3=P
4x-9 =P
<span>n = 5
The formula for the confidence interval (CI) is
CI = m ± z*d/sqrt(n)
where
CI = confidence interval
m = mean
z = z value in standard normal table for desired confidence
n = number of samples
Since we want a 95% confidence interval, we need to divide that in half to get
95/2 = 47.5
Looking up 0.475 in a standard normal table gives us a z value of 1.96
Since we want the margin of error to be ± 0.0001, we want the expression ± z*d/sqrt(n) to also be ± 0.0001. And to simplify things, we can omit the ± and use the formula
0.0001 = z*d/sqrt(n)
Substitute the value z that we looked up, and get
0.0001 = 1.96*d/sqrt(n)
Substitute the standard deviation that we were given and
0.0001 = 1.96*0.001/sqrt(n)
0.0001 = 0.00196/sqrt(n)
Solve for n
0.0001*sqrt(n) = 0.00196
sqrt(n) = 19.6
n = 4.427188724
Since you can't have a fractional value for n, then n should be at least 5 for a 95% confidence interval that the measured mean is within 0.0001 grams of the correct mass.</span>
Answer:
it is not always possible to find their intersection point for e.g take
y=x-1 it's inverse is y=x+1 so their is no intersection point but generally u find intersection points by equalizing their functions
f(x)=f^-1(x)
Answer:
a) 0.0082
b) 0.9987
c) 0.9192
d) 0.5000
e) 1
Step-by-step explanation:
The question is concerned with the mean of a sample.
From the central limit theorem we have the formula:
a) 
The area to the left of z=2.40 is 0.9918
The area to the right of z=2.40 is 1-0.9918=0.0082
b) 
The area to the left of z=3.00 is 0.9987
c) The z-value of 1200 is 0
The area to the left of 0 is 0.5
The area to the left of z=1.40 is 0.9192
The probability that the sample mean is between 1200 and 1214 is
d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000
e) 
The area to the left of z=-112.65 is 0.
The area to the right of z=-112.65 is 1-0=1
Answer:
is 1
Step-by-step explanation:
sorry if im wrong this is my first time doing this
