Answer:
9.48*
Step-by-step explanation:
This is a right triangle. The formula for solving the Hypotenuse, or the longest side of the right triangle is A^2 + B^2 = C^2. If we put the numbers from the problem into the formula this is what we get :
3^2 + 9^2 = C^2
9 + 81 = C^2
90 = C^2
9.48 = C
* This is rounded, the exact number is closer to 9.486832980505138. Your class should tell you what to round to.
W^2=80
W SQRT80
W=8.94 ANS.FOR THE WIDTH
L=3*8.94=26.83 ANS.FOR THE LENGTH
PROOF:
240=8.94*26.83
240=240
Answer:
"76°" is the appropriate solution.
Step-by-step explanation:
Please find attachment of the diagram according to the given query.
The given values are:
In ΔDEF,
f = 610 inches
e = 590
∠E = 70°
∠F = ?
By using the law of sines, we get
⇒ 
On substituting the values, we get
⇒ 
On applying cross multiplication, we get
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
now,
⇒ 
⇒ 
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

The draws can happen in either of the 2 ways in (1) and (2) above
therefore total probability E is given as
E =
For the yellow one to be from 1994, it implies that the event to be chosen is

Since the total probability is given as E=0.054
then 
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