Answer:
$1.00
Step-by-step explanation:
Total Budget = $35
Cake Spent = $18
Remaining Money = 35 - 18 = $17
Since, each balloon is worth $4, to find how many balloons he will get with remaining money ($17), we divide the the remaining money by price of each balloon:
17/4 = 4.25
We can't have fractional balloons, so Abit can get 4 balloons, MAXIMUM.
4 balloons cost = $4 per balloon * 4 = $16
So, from $17 if he spends $16 for balloons, he will have left:
$17 - $16 = $1.00
Answer:
$919.98
Step-by-step explanation:
To solve using a financial calculator do
N=8
I/Y=10
PMT=85
FV=1000
CMPT PV get 919.98
To do by hand find the present value of the interst payments
85*(1-(1/1.1)^8)
which is 453.4687
Find the present value of the final ballon payment
1000*(1/1.1)^8
which is 466.507
take the sum
466.507+453.4678= 919.98
Answer: 9)−3
/4 or -0.75
Step-by-step explanation:
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
There are
ways of selecting two of the six blocks at random. The probability that one of them contains an error is
So
has probability mass function
These are the only two cases since there is only one error known to exist in the code; any two blocks of code chosen at random must either contain the error or not.
The expected value of finding an error is then
