Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by

In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
![p = 0.4[\tex]So[tex]E = \frac{r}{p} = \frac{1}{0.4} = 2.5](https://tex.z-dn.net/?f=p%20%3D%200.4%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3ESo%3C%2Fp%3E%3Cp%3E%5Btex%5DE%20%3D%20%5Cfrac%7Br%7D%7Bp%7D%20%3D%20%5Cfrac%7B1%7D%7B0.4%7D%20%3D%202.5)
The expected number of systems inspected until the first failed unit is 2.5
Answer:
- a) P(x) = 32000*1.04^x
- b) $37435
- c) During year 7
Step-by-step explanation:
<u>Given</u>
- Initial pay = $32000
- Increase rate = 4%
a. <u>Formula</u>
b. Year 5 is after 4 years, so we are looking for the value of P(4)
- P(4) = $32000*1.04^4 = $37435
c. <u>P(x) = 40000, x = ?</u>
- 40000 = 32000*1.04^x
- 1.04^x = 40000/32000
- 1.04^x = 1.25
- log 1.04^x = log 1.25
- x = log 1.25 / log 1.04
- x = 5.69, this is 6 years after
The required number of the years is 6 + 1 = 7
First you have to make 7 a fraction, every fraction with the denominator 1 is a whole number, so 7 should be 7/1,
7/1×2/3
=14/3
=4 2/3
answer=4 2/3
hope this helps....
Answer: AB will be parallel to A'B'.
Step-by-step explanation: We know the definition of dilation about the centre. It is defined as the enlargement or shrinken of the original figure keeping the centre of dilation or the figure as fixed.
We are given ΔUVW and AB is perpendicular to UW. Now, if we dilate the triangle about the origin, then the triangle will either enlarge or shrink keeping the centre fixed.
Let us consider the enlarged triangle, ΔU'V'W' as shown in the attached figure. Also, line AB will move to the new position A'B'. We can clearly see that both the lines are parallel to each other.
Thus, the line segments AB and A'B' will be parallel too each other.