Answer:
Hi ,
Cube of a number :
_______________
For a given number x we define cube
of x = x × x × x , denoted by x^3.
A given Natural number is a perfect
Cube if it can be expressed as the
product of triplets of equal factors.
Now ,
Write given number as product of
prime .
8788 = 2 × 4394
= 2 × 2 × 2197
= 2 × 2 × 13 × 169
= 2 × 2 × 13 × 13 × 13
= 2 × 2 × ( 13 × 13 × 13 )
Here we have only triplet of equal
factors i.e 13
To make 8788 into perfect Cube we
have multiply with 2.
Now ,
2 × 8788 = ( 2 × 2 × 2 ) × ( 13 × 13 × 13 )
17576 = ( 2 × 13 )^3 = ( 26 )^3 perfect
Cube
I hope this will useful to you.
Solve whats in the parentheses first
5k+25+4<21+6k
5k+29<21+6k
plug in a number in k for both sides
5(1)+29<21+6(1)
5+29<21+6
34<27
5(2)+29<21+6(2)
10+29<21+12
39<33
so 5(k+5)+4 is greater than 21+6k
Answer:
-41
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.