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2 years ago

What should you do if the severity of risk is low and the frequency of the risk event occurring is high?

1 answer:

4 0

If the severity of risk is low and the frequency of the risk event occurring is high thanwe should Avoid the risk.

High Frequency/ High Severity- Risks are almost certain to occur and when they occur impact will be very high. In such a case it is best to use Avoidance as a risk management technique. If avoidance is not possible then prevention and insurance techniques can be considered. High frequency/ Low severity- This more serious risk and occurrence is high but the impact is low. Examples of such risks include workers’ injuries and shoplifting. A common way to manage this type of risk is through Prevention.

Low frequency/ High severity- The impact of these kinds of risks is very high and can bankrupt a business. Insurance is the best technique to manage these risks that have low loss frequency and high loss severity. Low frequency/ Low severity- Retaining and self-insuring the risk. Risk occurrence is low and impact is also very low. In most cases, the costs of managing them outweigh the cost of retaining them.

Learn more about risk frequency here:- brainly.com/question/254161
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Probably D. Exotic Species

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3 years ago

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Navistar Electric issued 1000 debenture bonds 2 years ago with a face value of $5,000 each and a bond interest rate of 15% per y

Natasha2012 [34]

Answer:

$7,081.25

Explanation:

Face value = 5000

Coupon = 15% paid annually. Semi annual payment = 750/2 = 375

Time to maturity = 18 years

Interest rate = 10% compounded semi-annually

P = 375(P|A, 5%, 36) + 5000(P|F, 5%, 36)

P = 375(16.58131488) + 5000(0.17265193)

P = 6217.99308 + 863.25965

P = 7081.25273

P = $7,081.25

So, the present worth of one bond today is $7,081.25

4 0

2 years ago

You establish a straddle on Fincorp using September call and put options with a strike price of $80. The call premium is $7.00 a

Nina [5.8K]

Answer: $15.50

Explanation:

From the question, we are informed that someone establish a straddle on Fincorp using September call and put options with a strike price of $80 and that the call premium is $7.00 and the put premium is $8.50.

The most that can be lose on this position will be the addition of the call premium and the put premium. This will be:

= $7.00 + $8.50

= $15.50

6 0

3 years ago

Which of the following is an advantage that shopping at a physical store has over shopping online?

krok68 [10]

Answer:

A. Personal help and interaction wider.

Explanation:

Due to the physical contact and getting to physically recognise each other, the familiarity increase the level of personal relationship.

5 0

3 years ago

A fast-food restaurant has determined that the chance a customer will order a soft drink is 0.90. The proba- bility that a custo

azamat

Answer:

(a) The probability that the order will include a soft drink and no fries is 0.45.

(b) The probability that the order will include a hamburger and fries is 0.48.

Explanation:

Let the events be denoted as follows:

S = an order of soft drink

H = an order of hamburger

F = an order of french fries.

Given:

P (S) = 0.90

P (H) = 0.60

P (F) = 0.50

(a)

It is provided that the event of ordering a soft drink and fries are independent.

If events A and B are independent then the probability of event (A ∩ B) is:

$P(A\cap B)=P(A)\times P(B)$

Compute the probability that the order will include a soft drink and no fries as follows:

$P(S\cap \bar F)=P(S)\times P(\bar F)\\=P(S)\times[1-P(F)]\\=0.90\times (1-0.50)\\=0.45$

Thus, the probability that the order will include a soft drink and no fries is 0.45.

(b)

It is provided that the conditional probability that a customer will order fries given that he/she has already ordered a hamburger as, P (F|H) = 0.80.

The conditional probability of an event B given another event A has already occurred is:

$P(B|A)=\frac{P(A\cap B}{P(A)}$

Compute the probability that the order will include a hamburger and fries as follows:

$P(F|H)=\frac{P(H\cap F)}{P(H)}\\P(H\cap F)=P(F|H)\times P(H)\\=0.80\times 0.60\\=0.48$

Thus, the probability that the order will include a hamburger and fries is 0.48.

6 0

3 years ago