Ab = (7x-6) → b = (7x-6)/a (1)
bc = (12-2x) → b = (12-2x)/c (2)
Since (1) = (2) → (7x-6)/a = (12-2x)/c OR a/c = (7x-6)/(12-2x) (3)
Multiply both numerators of (3) by the SQUARE of their respective denominators;
(a*c²)/c = (7x-6)(12-2x)²/(12-2x)
Now simplify:
ac = (7x-6)(12x-2x) or ac = -14x² + 96x - 72
do you mind telling me the question?
Answer:
Step-by-step explanation:
You are to make 5 assemblies.
Each assembly requires the use of 1 Type A bolt.
To make the 5 assemblies, you need 5 Type A bolts.
The container of bolts has a total of 60 bolts.
The focus - Type A bolts - is 20 out of this 60.
The probability of obtaining a Type A bolt at all, is 20/60, which is = 1/3
(A) What is the probability of taking the exact number of Type A bolts you need for your 5 assemblies, if you randomly take 10 bolts from the container?
- The exact number of Type A bolts you need for the 5 assemblies is 5
1/3 × 5/10 = 5/30 = 1/6 = 0.167
(B) What is the probability of taking/having less than 5 Type A bolts out of the randomly selected 10 bolts? The solution is to sum up the following:
1/3 × 4/10 = 0.133
1/3 × 3/10 = 0.1
1/3 × 2/10 = 0.067
1/3 × 1/10 = 0.033
1/3 × 0/10 = 0
TOTAL = 0.333
Answer: 24.7
cos 54° = x/42
⇔ x = 42. cos 54°
⇔ x ≈ 24.7
Step-by-step explanation:
