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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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Corey bought 112 feet of porch boards at the lumberyard. If each foot cost $1.59,how much change would he get back from $200?

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3 years ago

In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer

stira [4]

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = percentage of patients stop smoking when given sustained care.

So, Null Hypothesis, $H_0$ : p = 80% {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, $H_A$ : p $\neq$ 80% {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of patients who stop smoking when given sustained care = 80.2%

n = sample of patients = 192

So, the test statistics = $\frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }$

= 0.07

The value of z test statistics is 0.07.

Also, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.07) = 1 - P(Z $\leq$ 0.07)

= 1 - 0.52790 = 0.4721

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.

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