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2 years ago

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Joe bought a computer three years ago for $6570. He assumed the value of the computer would depreciate at a rate of 6 3\4% per m

onth. What is the value of the computer at the end of three years?

2 answers:

Valentin [98]2 years ago

7 0

Answer:

305?

Step-by-step explanation:

if wrong sorry font be angry

zimovet [89]2 years ago

5 0

530.791093416 or 530.79$

Step-by-step explanation:

6570-6.75^36=530.791093416 or 530.79$

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a. In the general education course requirement at a college, a student needs to choose one each from social sciences, humanities

Elenna [48]

Using the Fundamental Counting Theorem, there are 240 ways to choose general education courses from these 4 areas.

<h3>What is the Fundamental Counting Theorem?</h3>

It is a theorem that states that if there are n things, each with $n_1, n_2, \cdots, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

Considering the number of options for each course, the parameters are given as follows:

$n_1 = 5, n_2 = 4, n_3 = 4, n_4 = 3$ .

Hence the number of ways is given by:

N = 5 x 4 x 4 x 3 = 240.

More can be learned about the Fundamental Counting Theorem at brainly.com/question/24314866

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5 0

2 years ago

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich

Bess [88]

Answer:

a). k = 2.0794

b). $A_{t}=A_{0}(2^{3t} )$

c). 13107200 cells

d). rate of growth after 6 hours = 27255112

e). 4.76 hours

Step-by-step explanation:

From the formula of bacterial population,

$A_{t}=A_{0}e^{kt}$

Where $A_{t}$ = Bacterial population after time t

$A_{0}$ = Initial population

k = relative growth factor

t = duration or time

a). For $A_{t}=2\times 50=100$ cells

and $A_{0}=50$ cells

Time t = 20 minutes = $\frac{1}{3}$ hours

Now we plug in these values in the formula,

100 = $50(e^{\frac{k}{3}})$

$e^{\frac{k}{3}}=2$

By taking natural log on both the sides of the equation,

$\frac{k}{3}(lne)=ln(2)$

k = 3ln(2) = 2.0794

b). To get the expression we will plug in the value of k in the formula.

$A_{t}=A_{0}e^{kt}$

Since k = 3ln(2)

$A_{t}=A_{0}e^{3t(ln2)}$

Let y = $e^{3t(ln2)}$

By taking natural log on both the sides of the equation,

lny = $ln(e^{3t(ln2)} )$

lny = 3t(ln2)ln(e)

ln(y) = 3t(ln2)

ln(y) = $ln(2)^{3t}$

y = $2^{3t}$

Now our expression will be $A_{t}=A_{0}(2^{3t} )$

c). Number of cells after t = 6 hours

$A_{6}=50(2^{3\times 6} )$

$A_{6}=50(2^{18})$

$A_{6}=50\times (262144)=13107200$

d). We can get the rate of growth by finding derivative of the expression

$\frac{d}{dt}(A_{t})=\frac{d}{dt}[A_{0}(e^{kt}})]$

= $A_{0}[ke^{kt}]$

Now $\frac{d}{dt}(A_{6})=k.A_{6}$

= 2.0794×13107200

= 27255112

e) From the given formula,

$A_{t}=A_{0}(2^{3t} )$

$1000000=50(2^{3t} )$

$2^{3t}=20000$

3t(ln2) = ln(20000)

t = $\frac{ln(20000)}{3(ln2)}$

t = $\frac{9.90349}{2.07944}=4.76$ hours

3 0

3 years ago

Will mark brainiest please help me.

Licemer1 [7]

Answer:

46 sides in all

Step-by-step explanation:

2 pentgons would have 10 sides

5 hexagons would have 30 sides

2 triangles 6 sides

7 0

3 years ago

Read 2 more answers

The school talent show is 90 minutes long. The average rate of the show is about 10 acts every 30 minute. At this rate, how many

SIZIF [17.4K]

Answer:

C. 30

90/30=3

3x10= 30

SECOND PART

the second year would only have 25 acts

100/4=25

7 0

3 years ago

Read 2 more answers

Adrianne invested $2000 in an account at a 3.5% interest rate compounded annually. She made no deposits or withdrawals on the ac

Musya8 [376]

Answer:

$2,295$ $dollars$

Step-by-step explanation:

This is our equation that I could find within this question.

$2000(1+.035)^4$

$==> 2000(1.035)^4$

$==> 2000 * 1.1475$

$==> 2,295$

7 0

3 years ago