Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population mean of the length of engineering conferences.
The variable has a normal distribution.
The information given is:
n= 84
x[bar]= 3.94
δ= 1.28
The formula for the Confidence interval is:
x[bar]±
*(δ/n)
Lower bound(Lb): 3.698
Upper bound(Ub): 4.182
Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242
I hope it helps!
we know that
If line b is perpendicular to line a, and line c is perpendicular to line a,
then
line b and line c are parallel
and two lines parallel have the same slope
so
<u>Find the slope of the line b</u>
Let

The formula to calculate the slope between two points is equal to


substitute



therefore
<u>the answer is</u>
<u>the slope of the line c is</u>

To solve this problem, let us first lay out all the
factors of each number.
48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
56 : 1, 2, 4, 7, 8, 14, 28, 56
The greatest number of bouquets that can be made would be
equal to the greatest common factor of the two numbers. In this case it would
be 8.
Answer:
<span>8 bouquets</span>
Lets take Gregs weight as “x”. This means that Justins weight is x-15, and x/2 = (x-15)-75.
If we take that last equation, lets combine like terms:
x/2 = x - 15 - 75
x/2 = x - 90
Now multiply both sides by 2 to get rid of the fraction
x = 2x - 180
Subtract 2x from both sides
x - 2x = -180
-x = -180
x = 180 — this is Gregs weight
Justins weight is x-15, so 180-15, which is 165 pounds. Hope this helped.
Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be :
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are : 
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19