Ask question

Ask question

All categories

3 years ago

13

The company A produces products a, and the company B produces products b. It is known that 3 percent of companies that produce p

roducts a go bankrupt over the year. Plus, the probability for companies producing products b to go bankrupt over the year is 6.1. What is the probability (given in decimal number) that both companies are going bankrupt over the year?
NB! Give the answer as a decimal number and round it to two decimal places.

Advanced Placement (AP)

1 answer:

slava [35]3 years ago

4 0

Answer:

probability = 0.183 %

Explanation:

given data

produce products = 3 %

probability for producing products b = 6.1

solution

Both companies produce different products and the likelihood of bankruptcy varies depending on the product produced. So, the bankruptcy potential of A and B companies is independent.

we will multiply the probability of each company's bankruptcy and that will be

probability = P(A=bankrupt) × P(B=bankrupt)

probability = 3% × 6.1%

probability = 0.183 %

You might be interested in

True or false- Gneiss is formed. The mineral grains in granite are flattened under pressure.

Vladimir79 [104]

Answer:

The correct answer is True

5 0

3 years ago

The Food and Drug Administration (FDA), the Occupational Safety and Health

Musya8 [376]

Answer: Regulatory agencies

Explanation:

Regulatory agencies are referred to as the agencies that implement and also enforce the laws that are being enacted by the legislatures.

The Food and Drug Administration (FDA), the Occupational Safety and Health Administration (OSHA), and the Federal Trade Commission (FTC) are all

examples of regulatory agencies as they ensure and monitor businesses so that they'll comply with stated laws.

In a case whereby the agencies find a company which doesn't comply with the laws, a penalty is being given to such company which can be in the form of a fine or their license can be revoked.

6 0

2 years ago

Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the

KIM [24]

Answer:

$T_{2020} = 15$

Explanation:

Given

$T_1 = 10$

$T_2 = 20$

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

<em>This is proved as follows;</em>

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

$T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1$

<em>Assume n = 3</em>

$T_{3} = \frac{T_1 + T_2}{2}$

<em>Multiply both sides by 2</em>

$2 * T_{3} = \frac{T_1 + T_2}{2} * 2$

$2T_{3} = T_1 + T_2$

<em>Assume n = 4</em>

$T_{4} = \frac{T_1 + T_2 + T_3}{3}$

$T_{4} = \frac{(T_1 + T_2) + T_3}{3}$

Substitute $2T_{3} = T_1 + T_2$

$T_{4} = \frac{2T_3 + T_3}{3}$

$T_{4} = \frac{3T_3}{3}$

$T_{4} = T_3$

Assume n = 5

$T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}$

$T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}$

Substitute $2T_{3} = T_1 + T_2$ and $T_{4} = T_3$

$T_{5} = \frac{2T_3 + T_3 +T_3}{4}$

$T_{5} = \frac{4T_3}{4}$

$T_{5} = \frac{(5-1)T_3}{5-1}$

<em>Replace 5 with n</em>

$T_{n} = \frac{(n-1)T_3}{n-1}$

<em>(n-1) will definitely cancel out (n-1); So, we're left with</em>

$T_{n} = T_3$

Hence,

$T_{2020} = T_3$

Calculating $T_3$

$T_{3} = \frac{10 + 20}{2}$

$T_{3} = \frac{30}{2}$

$T_{3} = 15$

Recall that $T_{2020} = T_3$

$T_{2020} = 15$

3 0

3 years ago

All of the following are examples of a four-year college except

juin [17]

What are the the colleges they listed?

5 0

2 years ago

Ubiór i wygląd zewnętrzny Bernarda Zygiera ( syzyfowe prace)

Vinil7 [7]

Ubidor anf wihger scholden

5 0

2 years ago