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anastassius [24]

3 years ago

8

Assuming there are no leap years, what would the total 3-year cost be for a laptop that has an average lifespan of 10 years, whe

n you purchased it for $780 and the daily cost to use it is $0.36
A $394.20
B$565.92
C$78.00
D$S1174 20

1 answer:

Dmitrij [34]3 years ago

5 0

Answer:

Step-by-step explanation:

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Is this right? If not let me know and explain how if u can. <br><br><br> (No link <3)

vlabodo [156]

Answer:

9/35

Step-by-step explanation:

There are a total of 15 bulbs to choose from randomly. First a new bulb is chosen. Chance is 9/15 because 9 out of 15 are new.

Then there are only 14 bulbs left. Then a used bulb is chosen. Chance is 6/14 since 6 out of remaining 14 are used.

Total chance for this to happen is 9/15 * 6/14 = 54/210 = 9/35.

8 0

3 years ago

Sarah is a computer engineer and manager and works for a software company. She receives a

daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = $a$

Given that:

$a_4=129\\a_{10}=207$

Formula for $n^{th}$ term of an Arithmetic Progression is given as:

$a_n=a+(n-1)d$

Where $d$ will represent the number of projects increased every year.

and $n$ is the year number.

$a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)$

Subtracting (2) from (1):

$78 = 6d\\\Rightarrow d =13$

By equation (1):

$129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90$

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

$a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233$

Each project pays $500

Earnings in the twelfth year = 233 $\times$ 500 = $116500

Sum of an AP is given as:

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938$

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 $\times$ 1938 = $969000

8 0

3 years ago

3 + 3(k + 3) = 6(k - 2) +9

ANTONII [103]

Step-by-step explanation:

3 + 3(k + 3) = 6(k - 2) + 9

3 + 3k + 9 = 6k - 12 + 9

12 + 12 - 9 = 6k - 3k

15 = 3k

k = 15/3

k = 5

7 0

3 years ago

to communication companies offer calling plans with company X it cost $0.35 to connect and then $0.05 for each minute with compa

Mama L [17]

Answer:

X Company charges $(0.2+0.01n)\times 10^2 \ cents$ more than Company Y

Step-by-step explanation:

Given:

Cost of X company = $\$ 0.35+(\$0.05\times 1 \ minute)$

Cost of Y Company = $\$ 0.15+(\$0.04\times 1 \ minute)$

for n minutes

Cost of X company for n minutes = $\$ 0.35+(\$0.05\times n \ minute)$

Cost of Y Company for n minutes = $\$ 0.15+(\$0.04\times n \ minute)$

Difference for n minutes we get by Subtracting Cost of Y Company for n minutes by Cost of X company for n minutes

Difference for n minutes = $(\$ 0.2+\$ 0.01n)$ = $(0.2+0.01n)\times 10^2 \ cents$

X Company charges $(0.2+0.01n)\times 10^2 \ cents$ more than Company Y

6 0

3 years ago

Jessie read 1/3 of a book in the morning, and she read some more at night. By the end of the day, she still had 1/5 of the book

Mrac [35]

7/15 She read 7/15 at night

6 0

3 years ago

Read 2 more answers