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Mariana [72]
3 years ago
11

How much should Jenna save up to prepare for car repairs and maintenance? (A lump sum emergency fund just for car repair/mainten

ance that may come up in the future)
Mathematics
1 answer:
Veronika [31]3 years ago
8 0

Answer:

$500 per year.

Step-by-step explanation:

Car repairs and maintenance expense is based on the usage of car as well as its present condition. If the car is bought from showroom and is new then it will incur only small amount of expense as its maintenance but if the car is used by various owners then it is likely that maintenance expense can be huge. The average car repair expense is $500 to $800 per year depending on car condition. Since Jenna car is new she should keep the repair estimate to be minimum that is nearly $500 per year or $42 per month.

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VERNON TOSED A COIN 20 TIMES THE RESULTS WERE8 HEADS AND 12 TAILS
Oxana [17]

Answer:

8+12=20

Step-by-step explanation:

Is there another part to the question or something let me know.

7 0
2 years ago
Read 2 more answers
Help math plz! Giving brainlist :D
adoni [48]
No, there is not common ratio...

0.972  (common ratio is 0.3...10.8*.3*.3)

you can use the geometric mean to find this center value easily...

(18*8)^(1/2)=12
7 0
3 years ago
A factory makes 12 bikes in 3 hours.
zysi [14]
12/3 = x/8

Cross multiply

3x = 96

Eliminate

3x/3 = 96/3

x = 32
6 0
3 years ago
Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How
Scorpion4ik [409]

let's check how much is it after 2 years firstly.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275


Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

4 0
3 years ago
Find the product.<br><br> (r + s)2
mario62 [17]
When you've got a squares bracket, this simply becomes the bracket twice, therefore, (r+s)2 becomes (r+s)(r+s)
You've then go to use FOIL, and this gives you r^2+rs+sr+s^2, and this cannot be simplified any further.
Therefore, the product of this is r^2 + s^2 + rs + sr
Hope this is what you're looking for :)
3 0
3 years ago
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