0.29m+10=0.35m
0.29m-0.29m+10= 0.35m-0.29m
10=0.06 m
Divide by 0.06 for 10 and 0.06
10/0.06= 0.06/0.06 m
m= 166.66666
Answer is m = 166.66666- the number 6 continues
Given that In 3 years, Dianna will be 4 times as old as she was 33 years ago, her present age is 45.
<h3>How old is Dianna now?</h3>
Given that, in 3 years, Dianna will be 4 times as old as she was 33 years ago.
Let x represent the age of Dianna presently.
Dianna's age in 3 years time will be: x + 3
Dianna's age 33 years ago is: x - 33
Since in 3 years, she will be 4 times as old as she was 33 years ago.
x + 3 = 4( x - 33 )
We solve for x
x + 3 = 4x - 132
3 + 132 = 4x - x
135 = 3x
x = 135/3
x = 45
Given that In 3 years, Dianna will be 4 times as old as she was 33 years ago, her present age is 45.
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Answer:
Step-by-step explanation:
Total number of antenna is 15
Defective antenna is 3
The functional antenna is 15-3=12.
Now, if no two defectives are to be consecutive, then the spaces between the functional antennas must each contain at most one defective antenna.
So,
We line up the 13 good ones, and see where the bad one will fits in
__G __ G __ G __ G __ G __G __ G __ G __ G __ G __ G __ G __G __
Each of the places where there's a line is an available spot for one (and no more than one!) bad antenna.
Then,
There are 14 spot available for the defective and there are 3 defective, so the arrange will be combinational arrangement
ⁿCr= n!/(n-r)!r!
The number of arrangement is
14C3=14!/(14-3)!3!
14C3=14×13×12×11!/11!×3×2
14C3=14×13×12/6
14C3=364ways
Answer:
Where
and 
Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:



So then is appropiate use the normal distribution to find the probabilities for 
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter
is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: 
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and 
Since the distribution for X is normal then the distribution for the sample mean
is also normal and given by:



So then is appropiate use the normal distribution to find the probabilities for 