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yan [13]
1 year ago
5

If Ethan takes out a $5,000 loan to buy a motorcycle and his payments are $168.45 a month for 36 months, how much money will he

pay in interest over the life of the loan?
Mathematics
1 answer:
Vilka [71]1 year ago
4 0

Answer:

  $1064.20

Step-by-step explanation:

36 payments of $168.45 have a total of $6064.20. The excess over the loan amount of $5000 is the interest paid:

  $6064.20 -5000 = $1064.20 . . . . interest paid

You might be interested in
Linda commutes a total of 54 miles to and from work every day. She needs a new car, and good gas mileage is important to her. Sh
Alchen [17]
The answer is 1 gallon.

Miles per gallon(mpg) is computed by dividing the distance traveled by the how many gallons used. So you can derive a formula for how many gallons you would use given the mpg. You will end up with:

gallon =  \frac{miles travelled}{miles per gallon}

The problem asks for how many gallons of gas she will safe in a five-day work work week. So first you need to compute how many miles that would be. 

54 miles/day x 5days =  270 miles

So in a five day work week, she will travel 270 miles.

Now to see how much gas she will save, compute how many gallons she will use up for each car, given the mpg of each and find the difference. 

First model:30 mpg

gallons = \frac{270miles}{30mpg}
gallons = 9g
 
This means that with the first model, she will have used up 9 gallons in a 5-day work week.

Second model: 27 mpg
g = \frac{270mi}{27mi/g}
g = 10g

This means that with the second model, she will have used up 10g in a 5-day work week. 

Now for the last bit. How much will she save? You can get that by getting the difference of how many gallons each car would have used up.

10gallons - 9gallons = 1g

So she would have saved 1 gallon of gas if she buys the first car instead of the second. 
 
8 0
1 year ago
Read 2 more answers
Steve stayed after school for extra activities 55% of the 180 school days this year. How many days did Steve stay after school?
musickatia [10]

Answer:

11/20 or 99/180 days

Step-by-step explanation:

55/100 converted to ?/180

180/100=1.8

so you would do the same on the numerator which is 55*1.8=99

99/180= 33/60=11/20

7 0
1 year ago
While vacationing, Rosalie sees a mountain and its reflection on the still water of a lake. She draws the mountain and its refle
Greeley [361]

Answer:

I would say (2,-9) is her projection

6 0
2 years ago
A meter is 3.37 inches more than a yard swimmer a swims 100 meters and swimmer b swims 100 yeards how many more feet did swimmer
olga_2 [115]
Swimmer a swims 100 meters, which is 100 1-meters, which is 100
(1 yards +3.37 inches) = 100 yards + 337 inches.

1 yard is 3 feet, so 100 yards are 300 feet.

100 in is 8.33 feet so
337 in is (337*8.33)/100=28.07 feet


Swimmer b swims 100 yards, which is 300 feet

Swimmer a swam 28.07 feet.


4 0
2 years ago
A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr
erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034

The estimated standard error is:

\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0
2 years ago
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