The number in the ones place= X
The number in the tenths place= y
y+x=14.
y=14-x
Original placement of numbers=10y+x
New placement of numbers(after subtraction of 36)= 10x+y
10y+ x-36=10x+y
Substitute y with 14-x
So we would get
10(14-x)+ x-36=10x+(14-x)
=140-10x+x-36=10x+14-x
After transposing we get
140-14-36=10x-x+10x-x
=90=18x
X=90/18
x=5
Y=14-x
=14-5
=9
So the two digit number is
10y+x
=90+5
=95
so the investigator found the skid marks were 75 feet long hmmm what speed will that be?
![s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B30fd%7D~~%20%5Cbegin%7Bcases%7D%20f%3D%5Cstackrel%7Bfriction%7D%7Bfactor%7D%5C%5C%20d%3D%5Cstackrel%7Bskid%7D%7Bfeet%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20f%3D%5Cstackrel%7Bdry~day%7D%7B0.7%7D%5C%5C%20d%3D75%20%5Cend%7Bcases%7D%5Cimplies%20s%3D%5Csqrt%7B30%280.7%29%2875%29%7D%5Cimplies%20s%5Capprox%2039.69~%5Cfrac%7Bm%7D%7Bh%7D)
nope, the analysis shows that Charlie was going faster than 35 m/h.
now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been
![s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B30fd%7D~~%20%5Cbegin%7Bcases%7D%20f%3D%5Cstackrel%7Bfriction%7D%7Bfactor%7D%5C%5C%20d%3D%5Cstackrel%7Bskid%7D%7Bfeet%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20f%3D%5Cstackrel%7Bdry~day%7D%7B0.7%7D%5C%5C%20s%3D35%20%5Cend%7Bcases%7D%5Cimplies%2035%3D%5Csqrt%7B30%280.7%29d%7D%20%5C%5C%5C%5C%5C%5C%2035%5E2%3D30%280.7%29d%5Cimplies%20%5Ccfrac%7B35%5E2%7D%7B30%280.7%29%7D%3Dd%5Cimplies%2058~ft%5Capprox%20d)
It is easier to do if we get rid of the decimals, so lets multiply both numbers by 10:
5.8/1.2 = 58/12
and reduce the fraction:
= 29/6
then operate:
= 4.83
Which data set has a median of 15? 9, 17, 13, 15, 16, 8, 12 18, 15, 11, 14, 19, 15, 6 7, 16, 14, 16, 11, 7, 17 18, 9, 19, 16, 6,
ivanzaharov [21]
Answer:
The middle number of the data set or average of the two middle numbers in even numbered sets.
Step-by-step explanation:
A median is the middle point of a data set. We order the numbers from least to greatest and find the number directly in the middle of the list. If there are an even number in the set, then we take an average of the middle two.
The data sets are not separated. However, in the sets you have order them least to greatest each. Then count in from both sides to the middle. That number is the middle.
The weight of a plane with 81 gallons of fuel in the tank is about 44 tons
