There are 18 coins total in her bag:
7+3+8=18coins
3 of the 18 coins are dimes, so the odds of first drawing a dime are 3/18
P(dime)=3/18
Since the nickel was taken out and not replaced, there are now 17 coins in her purse.
18-1=17coins
8 of the 17 remaining coins are nickels, so the odds of drawing a nickel are 8/17.
P(nickel)=8/17
Now we multiply the two probabilities by eachother to find out the odds of both occurring...
P(dime)*P(nickel)=
3/18*8/17=24/306=8/103=0.078=8%
The probability of grabbing a dime and then a nickel are 8/103 or 0.078 or an 8% chance, whichever way you'd prefer to write it.
Answer:
see below
Step-by-step explanation:
4 ^(3x-1)=3
Take the natural log of each side
ln(4 ^(3x-1))= ln (3)
We know that ln ( a^b) = b ln a
(3x-1) ln 4 = ln 3
Divide each side by ln 4
3x-1 = ln(3)/ (ln4)
Add 1 to each side
3x = ln(3)/ ln (4) +1
Divide by 3
x =1/3 [( ln 3)/ ln (4)] + 1/3
x = .792461
Answer:
2n+2
Step-by-step explanation:
-n + (-3) +3n + 5 -3
(-3+5) =2
-n +2+n+3n
3n-n = 2n
2n+2
In the figure below
1) Using the theorem of similar triangles (ΔBXY and ΔBAC),
Where
Thus,
thus, BC = 7.5
2) BX = 9, BA = 15, BY = 15, YC = y
In the above diagram,
Thus, from the theorem of similar triangles,
solving for y, we have
thus, YC = 10.
