F(x) = x² + 4x + 10
f(x) - 10 = x² + 4x
perfect square:
x² + 4x + 4 ⇒ (x + 2)²
(x + 2)² = x(x+2) + 2(x+2) = x² + 2x + 2x + 4 = x² + 4x + 4
f(x) - 10 + 4 = x² + 4x + 4
f(x) - 6 = (x+2)²
f(x) = (x+2)² + 6
Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
Answer:
The position and orientation of the new angle may vary, but it should have the same angle measure as ∠BAC
.Step-by-step explanation:
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.