Ronaldo has 50/250 BI liability insurance. He loses control of his car and injures 18 children

Mathematics

1 answer:

Alex73 [517]3 years ago

8 0

Answer:

The insurance company will pay $250,000 in total for this lawsuit.

Ronaldo will be personally responsible for $110,000.

Explanation:

BI stands for bodily injury.

Bodily Injury liability insurance covers the damages caused to third parties (other drivers or persons not including you).

In the format 50/250, the first number is the amount covered for one person in the accident, and the second number is the limit that the insurance company covers in the same accident.

Since each child is awarded $20,000 as a result of a lawsuit, the total legal responsability is 18 × $20,000 = $360,000.

Hence, the limit of $ 250,000 covered by the insurance company is exceeded meaning that the insurance company will pay the limit of $250,000.

Therefore, Ronaldo will be personally responsible for the difference: $360,000 - $250,000 = $110,000.

You might be interested in

• If the new image is smaller than the old image would

mars1129 [50]

Answer:

Less than one

Step-by-step explanation:

5 0

2 years ago

What is the value of x?

Bogdan [553]

Answer:

Step-by-step explanation:

90+15=105

105÷5=21

6 0

2 years ago

In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th

marysya [2.9K]

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}$

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

First we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

Next we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$