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3 years ago

Click on all items which will cause a decrease in the cost of an auto insurance policy.

Mathematics

2 answers:

qwelly [4]3 years ago

8 0

Answer:

Auto insurance is an insurance that protects you against financial loss in case of an accident or auto theft. To avail the insurance, you have to pay premiums based on interest rates.

The items that can cause a decrease in the cost of an auto insurance policy are :

You are a mature driver. (younger drivers are given high interest rates)

You have had driver's training.(Professionally trained people have very little chance of causing any accident.)

You have multi-vehicle coverage on your policy.(insurance companies give discounts if more than one vehicle is insured under them)

ololo11 [35]3 years ago

6 0

The answers for this question are 2 , 3 , and 5. these 3 make the most sense.

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Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

3 years ago

Let x stand for the number of bananas in one bag. write an equation to find x.

xz_007 [3.2K]

(7-1)/3=x
(14-2)/4=x
1. x=2
2. x=3

If you are looking for the number of bananas per bag you must first subtract your remainders from the total number of bananas, then divide those bananas by the number of bags and you have your answer.

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Naily [24]

The median is 44 (the center line)

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olasank [31]

${m}^{2} - n \\ Given, \: \: m = 3 \: \: n = 2, then, \\ \: \: {(3)}^{2} - 2 \\ = 9 - 2 \\ = 7$