30kx-6kx=8
24kx=8
x=8/(24k)
x=1/(3k)
so the answer should actually be D
Correct question is;
A bag contains 10 counters. 6 of them are white. A counter is taken at random and not replaced. A second counter is taken out of the bag at random. Calculate the probability that only one of the two counters are white
Answer:
probability that only one of the two counters is white = 8/15
Step-by-step explanation:
To solve this question, first of all, let's look at probability we would have to either draw two white counters or two non-white counters (4/10 * 3/9).
Probability(draw 2 white counters) = (6/10 × 5/9) = 30/90 = 1/3
Probability(draw 2 non-white counters) = (4/10 × 3/9) = 2/15
Now, In all other cases, we'll draw exactly one white and one non-white counter, so the odds of this would be;
P(one white counter and one non-white counter) = 1 - [1/3 + 2/15)
= 1 - 7/15 = 8/15
Answer: 2 to the power of 0 can be expressed as 20
Step-by-step explanation: tbf im not sure, explain clearly
We have that
1)
cot 70=x/12
Cot
70=[opposite side]/[adyacent side]=x/12
The
option 1 is correct
2)
cot 70=12/x
Cot
70=[opposite side]/[adyacent side]=x/12
The
option 2 is not correct
3)sec 20=x/12
Sec 20
=1/cos20
Cos
20==[opposite side]/[hypotenuse]=12/z
1/(1/cos20)------->
1/(12/z)------> z/12
The
option 3 is not correct
<span>
4) sec 20=12/x</span>
Sec
20 =1/cos20
Cos
20==[opposite side]/[hypotenuse]=12/z
1/(1/cos20)------->
1/(12/z)------> z/12
<span>The
option 4 is not correct</span>
the correct value is
1) cot 70=x/12
Ans
(2x-1/x)^5
=-48x^3+80x-40/x^2+10/x^3-1/x^5
