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eimsori [14]
4 years ago
8

Tina wants to save money for school. Tina invests $1,100 in an account that pays an interest rate of 7.25%. How many years will

it take for the account to reach $6,600? Round your answer to the nearest hundredth.
Mathematics
1 answer:
svlad2 [7]4 years ago
7 0
The amount of Tina money can be expressed in an exponent function like this:
an= $1100(1.0725)^n
The variable an represent the total money and variable n is the years needed to achieve that amount.

Then, the time needed for the money to reach $6,600 would be:

an= $1100(1.0725)^n
$6,600= $1100(1.0725)^n
$1,100(6)= $1100(1.0725)^n
6= (1.0725)^n
n= log1.075 6
n= 24.78
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George invests $5,000.00 in a savings account which pays 7% compounded continuously. Consider the following formula, where A is
vovikov84 [41]

Answer:

it's D: 7087.76

Step-by-step explanation:

since you already write the formulae, all you have to do it's just calculate with that

5 0
3 years ago
-(28-4x)=92 This is a two-step equation with distributive property
djverab [1.8K]
-(28-4x)=92
-24+4x=92
4x=92+24
4x=116
x=29


X=29
4 0
3 years ago
Express 0.9534 as a fraction
Solnce55 [7]
Express 0.9534 as a fraction

<span>0.9534 × 10 × 10 × 10 × 10 = 9534
</span><span>1 × 10 × 10 × 10 × 10 = 10000
</span>9534/10000

Divide by GCF:-

GCF = 2

9534 ÷ 2 = 4767
1000 ÷ 2 = 500
4767/500
^^^Improper fraction
Convert to mixed number:-

4767 ÷ 500 = <span>9.534

500 × 9 = 4500</span>

<span>4767 - 4500 =  267</span>

<span>9 = whole number</span>
<span>267 = </span>numerator
<span>500 = </span>denominator
<span>
9 267/500
 
0.9534 = 9537/10000 = 4767/1000 = 9 267/500 </span><span />
6 0
3 years ago
Evaluate 5(6) -c ifc=7
ahrayia [7]
23.

Work:
5(6)-7
5(6)=30
30-7=23
5 0
3 years ago
Read 2 more answers
According to astronomers, what is a "light year" answer key
liberstina [14]
A light-year is defined as the distance the light can cover in one year.

The light travels at speed of c=3 \cdot 10^8 m/s, and one year converted into seconds is
1 y = 365  \frac{d}{y} \cdot 24  \frac{h}{d} \cdot  60 \frac{m}{h} \cdot 60  \frac{s}{m}=3.15 \cdot 10^7 s

So the distance covered by the light in one year is
S=vt =(3 \cdot 10^8 m/s)(3.15 \cdot 10^7 s)=9.45 \cdot 10^{15} m
And this is the distance that corresponds to one light-year.
3 0
3 years ago
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