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Natasha2012 [34]
2 years ago
7

You have just spoken to your insurance agent and you are interested in investing in a 20-Payment Life insurance policy. Given th

at you are a 25-year-old healthy female, determine the annual premium for a policy with a face value of $55,000.Use the table provided below and round your answer to the nearest cent where necessary.
a.
$1,377.20
c.
$1,427.80
b.
$790.90
d.
$1,475.65

Mathematics
2 answers:
lakkis [162]2 years ago
8 0

Answer:

a.  $1,377.20

Step-by-step explanation:

To determine the annual premium for the policy, you have to divide the face value by 1,000:

$55,000/1,000= $55

Then, you have to multiply this result for the rate found on the table, which according to the information given is $25.04:

$55*$25.04= $1,377.20

The annual premium for the policy is $1,377.20.

Lesechka [4]2 years ago
6 0

Answer: You can use this to help you out:

Determine the term life insurance amount per thousand on a 20-year-old male for a 10-year policy given that the face value of the policy is $65,000 and the annual premium is $291.85.

a.

$0.00449

c.

$0.2227

b.

$4.49

d.

$222.72

First find how many thousands on the amount of the face value

You do that by dividing the amount of the face value by 1000

65,000÷1,000=65

So the term life insurance amount per thousand is

291.85÷65=4.49...answer

Read more on Brainly.com - brainly.com/question/4241979#readmore

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Basile [38]

Constraint 1:

Let the total number of running shoes be = R

Let the total number of leather boots be = L

As the given number of total shoes are 48,

The equations becomes,

R + L = 48............(1)

Constraint 2:

As running shoes are twice the leather boots, equation becomes,

R = 2L..............(2)

Putting the value of R from equation(2) in equation (1)

2L+ L=48

3L=48

L=16

Now putting the value of L in equation(2)

R= 2L

R = 2*16

R=32

Hence, Amanda needs 16 pairs of leather boots and 32 pairs of running shoes.

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2 years ago
What is 10 times 7 thousands in unit form
Irina18 [472]
10 times 7 is 70.
So, it is 70 thousands.
Or in other words, seven ten-thousands.
or just write it normally: 10*7*1000=70000=seventy thousand
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Answer:

△RST ~ △RYX by the SSS similarity theorem. Which ratio is also equal to RT/RX and RS/RY ?

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Step-by-step explanation:

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A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st
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Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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