Answer:
Consider the following calculations
Step-by-step explanation:
Since 1 Blimp uses 2 components of B and C each
=> choosing 2 components of B(remaining after using in other prototypes) for 1st model= 22C2
choosing 2 components of B(remaining after using in other prototypes) for 2nd model= 21C2
choosing 2 components of B(remaining after using in other prototypes) for 3rd model= 20C2
choosing 2 components of B(remaining after using in other prototypes) for 4th model= 19C2
choosing 2 components of B(remaining after using in other prototypes) for 5th model= 18C2
and choosing 2 components of C(remaining after using in other prototypes) = 24C2
Similarly for C
P(5 prototypes of Blimp created)=[(22C2 / 25C2 )*(24C2 / 25C2 )] + [(21C2 / 25C2 )*(23C2 / 25C2 )]+[(20C2 / 25C2 )*(22C2 / 25C2 )]+[(19C2 / 25C2 )*(21C2 / 25C2 )]+[(18C2 / 25C2 )*(20C2 / 25C2 )]
B- 1.23 You can do that whit an app name photomath :D
Answer:
It's 5.
Step-by-step explanation: Please click the crown thing near my answer if correct!
Using the Pythagorean theorm,
a^2 + b^2 = C^2, where a nd b are the side lengths and c is the hypotenuse ( diagonal line)
If it was a square both sides would be the same length so we would have:
13^2 +13^2 = c^2
169 + 169 = c^2
338 = c^2
c = SQRT(338) = 18.38
18.38 is larger than the 15.26 given in the problem , so the room is not a square.
4(4) + 7(3) + 3(4) - 2(3)
(16 + 21) + (12 - 6)
37 + 6
43
Your answer is B. 43