Answer:
The expected number of defective batteries to be pulled out is 0.9, which rounded to the nearest integer gives a total of 1, that is, 1 of the 3 batteries is expected to be defective.
Step-by-step explanation:
Given that a box contains 3 defective batteries and 7 good ones, and I reach in and pull out three batteries, to determine what is the expected number of defective batteries, the following calculation must be performed:
3 + 7 = 100
3 = X
10 = 100
3 = X
3 x 100/10 = X
300/10 = X
30 = X
3 x 3/10 = X
0.9 = X
Therefore, the expected number of defective batteries to be pulled out is 0.9, which rounded to the nearest integer gives a total of 1, that is, 1 of the 3 batteries is expected to be defective.
Answer:
2 touchdowns, 3 field goals
Step-by-step explanation:
The number of touchdowns cannot be more than 3, so it is relatively easy to find the solution by trial and error.
23 is not divisible by 3, so 0 touchdowns is not a solution
23 -7 = 16 is not divisible by 3, so 1 touchdown is not a solution
23 -14 = 9 is divisible by 3, so 2 touchdowns and 3 field goals is a solution
21 -21 = 2 is not divisible by 3, so there is only one solution.
Answer:
Assuming you mean 2+ (4/5), no
Step-by-step explanation:
=2* (5/5) + (4/5)
=10/5 + 4/5
=14/5
if you mean (2+4)/5,
its 6/5, so still not 1/3
Try this option:
6) similar: ∠C=∠E=60; ∠ABC=∠DBE.
7) similar:
ΔLMN: ∠L=75; ∠M=50; ∠N=180-125=55°; ΔPOQ: ∠O=75; ∠Q=55; ∠P=180-130=50°.
