<em><u>Answer 1: 177 – 77 = 100 </u></em>
<em><u>Answer 2: (7+7) * (7 + (1/7)) = 100</u></em>
Answer:
0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.
The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.
Step-by-step explanation:
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.
First six not defective, each with 0.98 probability.
7th defective, with 0.02 probability. So

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.
Find the expected number and variance of the number of components tested before a defective component is found.
Inverse binomial distribution, with 
Expected number before 1 defective(n = 1). So

Variance is:

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.
Answer:
∠ACB = 28.5
Step-by-step explanation:
In triangle ABC,
angle CAB = x-3
angle ABC = 4X-3
It is also given that AB=CB
We know that if two sides of any triangle are equal then corresponding sides must also be equal.
Hence
angle ACB = angle CAB
angle ACB = x-3 {Given that angle CAB = x-3 }
We know that sum of all three angles of a triangle is 180 degree so let's add given angles
angle ACB + angle CAB + angle ABC = 180
(x-3) + (x-3) + (4x-3) = 180
x-3 + x-3 + 4x-3 = 180
6x-9 = 180
6x = 180+9
6x = 189
x = 189/9
x= 31.5
Now plug value of x into
angle ACB = x-3 = 31.5-3 = 28.5
Hence final answer is
∠ACB = 28.5
Answer:
Correct option is
B
90
∘
,90
∘
,90
∘
Let AB and CD be two lines Intersecting at O, such that, ∠AOD=90
∘
Now, ∠AOD=∠COB=90
∘
(Vertically opposite angles)
⟹∠AOD+∠DOB=180
o
(Angles on a straight line)
⟹90+∠DOB=180
o
∠DOB=90
∘
∠DOB=∠AOC=90
∘
(Vertically opposite angles)
Thus, all angles are 90
∘
.