Answer:
Mean= 
=13.28 ≈13
For median arrange in ascending or descending order. i am arranging in ascending order
9,11,11,13,15,17,17
cancel from left and right until one number is left
median=13
<span>V(o) = 1/3 πr²h - old volume
</span>V(n) = 1/3 π(2r)²(h/2)=1/3 π(2r)²(h/2)=1/3*4/2* πr²h=1/3*2* πr²h
V(n)/V(o)=1/3*2* πr²h/1/3 πr²h=2/1
V(n)/V(o)=2/1 - ratio
the volume will be doubled
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for


which indeed gives the recurrence you found,

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that

, and substituting this into the recurrence, you find that

for all

.
Next, the linear term tells you that

, or

.
Now, if

is the first term in the sequence, then by the recurrence you have



and so on, such that

for all

.
Finally, the quadratic term gives

, or

. Then by the recurrence,




and so on, such that

for all

.
Now, the solution was proposed to be

so the general solution would be


Answer:
2/5=0.4h
Step-by-step explanation:
Use V=X/t
t=X/V
=50/125=2/5=0.4h
Answer:
Yes
Step-by-step explanation:
A full can of tuna has a certain amount in it, if a portion is removed from the can it is therefore subtracted from the total amount. This means that Joey's work is correct and the tuna can would have
5.5 oz - 2.88 oz = 2.62 oz
The best way to tell if his work is correct would be to weigh the remaining tuna in the can and see if it matches Joey's answer.