Answer:
x=7
7-4=3
7+2=9
Step-by-step explanation:
Triangle ABE and Triangle CDB are similar
4*3=12
ABE is 3 times bigger than CDB
3(x-4)=x+2
3x-12=x+2
2x-12=2
2x=14
x=7
7-4=3
7+2=9
Answer:
It is the same answer, Because b is the same as a but in fraction form, and c is the same as b but in improper fraction form, so that means they are the same answer
Step-by-step explanation:
Answer:
Step-by-step explanation:
This problem is all about bases and exponents. Because we have a quotient and the bases are both 5's, that means that we can use the rule of exponents for quotients to rewrite and simplify:
That's the simplification as long as you are "allowed" to leave the exponent as a negative number.
Answer:
A
Step-by-step explanation:
SAS = side-angle-side
This means that, in order to prove that the triangles are congruent, they must have two congruent sides with the angle between them to the same.
We know that sides AB, ED, AC, and DF are all congruent as they all have a single mark through them. From this, you can conclude that the triangles already share two sides. All we need now is the angles in between to be congruent. This means that angle A and angle D need to be congruent.
I hope this helps!
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.