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Stella [2.4K]
3 years ago
7

4. A typical smartphone battery (yes that includes iphones) degrades at approximately 1%

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
5 0

Answer:

a. f(t) = (0.99)^{t} 100%\\

b. The x - intercept represents the time at which the battery life drops to 0 %.

c. The y - intercept represents the initial battery life.

Step-by-step explanation:

a. Construct an equation using f(t) and t that represents the relationship between

"f(t)" the battery capacity of a typical smartphone after "t" months and "L" the

number of months the smartphone has been in normal use.

Let the initial battery life be I.

After one month, it decreases by 1%. So, our new value, I' = initial value - decrease = I - 1%I = 99%I.

After two months, it decreases by 1% to our new value I". So, our new value, I" = initial value - decrease = I' - 1%I' = 99%I' = 99%(99%)I = (99%)²I

After three months, it decreases by 1% to our new value I'". So, our new value, I"' = initial value - decrease = I" - 1%I" = 99%I" = 99%(99%)²I = (99%)³I

We see a pattern here.

After t months, f(t)

f(t) = (0.99)^{t} 100%\\

b. What does the horizontal intercept represent in context? (The x-intercept)

The x - intercept represents the time at which the battery life drops to 0 %.

c. What does the vertical intercept represent in context? (The y-intercept)

The y - intercept represents the initial battery life.

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puteri [66]

Answer:

Im not sure about this but maybe its $10.15

Step-by-step explanation:

8.25 x 5= 41.25

92- 41.25=50.75

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7 0
3 years ago
How many positive integers between 1000 and 9999 inclusive
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Answer:

Step-by-step explanation:

a.

first number is  1000-1+9=1008

9)1000(1

    9

-------

     10

       9

    -----

       10

         9

       ----

         1

       ----

last number is 9999

9| 9999

  ---------

    1111 |0

    --------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

b.

first digit=1000

last digit=9999-1=9998

2 |9999

  ---------

  |4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

d.

total  numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

e.

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

  ---------

  |  142-6

1000-6+7=1001

7 | 9999

  |---------

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9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

        70

       ----

        300

        280

        ------

           20

           ---

1000-20+35=1015

35)9999(285

     70

    ----

     299

     280

     -----

        199

        175

        ----

          24

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last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

7 0
3 years ago
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IrinaVladis [17]

Answer:

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Step-by-step explanation:

Well 9 - 4 1/12 is 4 11/12

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4 0
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