Answer:
$7520.55
Step-by-step explanation:
Cost of truck = $8000
Residual value = $1000
Estimated useful life = 7 years
Depreciation = (cost of asset - salvage value) / useful life
Depreciation = (8000 - 1000) / 7
Depreciation = 7000 / 7
Depreciation = $1000
Truck was purchased on July 9, Therefore, Depreciation by the end of year one will be;
Number of days between July 9 and year end = 175 days
Daily Depreciation = $1000 / 365 = $2.739
Total Depreciation by year end = (daily Depreciation * 175 days) = $479.45.
Book value at the end of year 1 = (8000 - 479.45)
= $7520.55
Answer:
6 servings
Step-by-step explanation:
To find the number of servings, divide the total amount by the quantity per serving.
You can compute this using a calculator:
5.333/0.875 ≈ 6.095 ≈ 6 servings
__
Or, you can do the mixed-number math:
(5 1/3)/(7/8) = (16/3)(8/7) = 128/21 = 6 2/21 ≈ 6
There were 6 servings in the container.
The circumference of a circle is pi(d) so if you multiply pi(4) = 3.141592654(4) you would get 12.5663706144
Answer:
The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%
Step-by-step explanation:
Given
Before 1995
(Br) Brown = 30%
(Y) Yellow = 20% =0.2
(R) Red = 20%
(G) Green =10% =0.1
(O) Orange = 10%
(T) Tan = 10%
After 1995
(Br) Brown = 13%
(Y) Yellow = 14% =0.14
(R) Red = 13%
(G) Green = 20%
= 0.2
(O) Orange = 16%
(Bl) Blue = 24%
Since there are two bags, let A be the bag from 1994, and B be the bag from 1996
Then let AY imply we drew a yellow M&M from the 1994 bag
AG implies we drew a green M&M from the 1994 bag
BY implies imply we drew a yellow M&M from the 1996 bag
BG implies we drew a green M&M from the 1996 bag
P(AY) =0.2
P (BY) = 0.14
P(AG) =0.1
P(BG) =0.2
Since the draws from the 1994 and 1996 bag are independent,
therefore
![P(AY n BG) = 0.2 * 0.2 = 0.04 -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014 --------(2)\\](https://tex.z-dn.net/?f=P%28AY%20n%20BG%29%20%3D%200.2%20%2A%200.2%20%3D%200.04%20%20-------%281%29%5C%5CP%28AG%20n%20BY%29%20%3D0.1%20%2A%200.14%20%3D0.014%20%20%20--------%282%29%5C%5C)
The draws can happen in either of the 2 ways in (1) and (2) above
therefore total probability E is given as
E =![P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54](https://tex.z-dn.net/?f=P%28%20AY%20n%20BG%29%20u%20P%28AG%20n%20BY%29%5C%5C%3D0.04%20%2B%200.014%20%3D0.O54)
For the yellow one to be from 1994, it implies that the event to be chosen is
![P(AYnBG) = 0.2*0.2](https://tex.z-dn.net/?f=P%28AYnBG%29%20%3D%200.2%2A0.2)
Since the total probability is given as E=0.054
then ![P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407](https://tex.z-dn.net/?f=P%28%28AYnBG%29%20%2FE%29%20%3D%5Cfrac%7B0.04%7D%7B0.054%7D%20%3D%200.07407)
Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E
Answer:
C.Segment AC=Segment A''C'' over 2
Step-by-step explanation:
We are given that
Dilation scale factor=2
A=(-3,3)
B=(1,-3)
C=(-3,-3)
We know that in dilation the size of figure change
Ratio of length of new segment to the length of segment of original figure=Scale factor
![\frac{A''C''}{AC}=2](https://tex.z-dn.net/?f=%5Cfrac%7BA%27%27C%27%27%7D%7BAC%7D%3D2)
![AC=\frac{A''C"}{2}](https://tex.z-dn.net/?f=AC%3D%5Cfrac%7BA%27%27C%22%7D%7B2%7D)
Option C is true.
C.Segment AC=Segment A''C'' over 2