A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 5% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 4% during the current year. In addition, it estimates that 15% of employees who had lost-time accidents last year will experience a lost-time accident during the current year. a. What percentage of the employee will experience a lost-time accident in both years (to 1 decimal)?

The percentage of the employee will experience a lost-time accident in both years is 0.0%

Step-by-step explanation:

Let A denote events that employees suffered lost-time accidents during the last year

Let B denote events that employees suffered lost-time accidents during the current year

P(A) = 5% = 0.05

P(B) = 4% = 0.04

P(B | A) = 15% = 0.15

(a) P (A ∩ B) = P(B | A) × P(A)

= 0.15 × 0.05

= 0.0075

= 0.0 (1 decimal place)

The probability that an employee will experience a lost- time accident in both years is 0.0

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.