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Pani-rosa [81]
2 years ago
14

Don’t know how to solve.

Mathematics
1 answer:
iVinArrow [24]2 years ago
3 0

Answer:

$122.88

Step-by-step explanation:

the phone decreases by 20% each year , that is

(100 - 20)% = 80% = \frac{80}{100} = 0.8

the phone reduces by a factor of 0.8 each year , then after 4 years

value = $300 × (0.8)^{4} = $122.88

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This property says you can rewrite 3x(4+5) as (3x4) + (3x5)
daser333 [38]
Distributive property
Step-by-step explanation:
Distributive property says that x(y + z) = (x*y) + (x *z)
6 0
3 years ago
Is 56 a multiple of 6?
andreyandreev [35.5K]

Good evening ,

_______

Answer:

No

____________________

Step-by-step explanation:

A number is a multiple of 6 if it can be divided by 6 .

Since 3 don’t divide 56 because 5 + 6=11 and 3 don’t divided 11 ,

Then 6 don’t divide 56

then , 56 isn’t a multiple of 6.

:)

3 0
3 years ago
At 12 noon in anchorage Alaska Janice noticed that the temperature outside was 12
Alisiya [41]
I assume that you want to know the temperature at a certain timing.
Well, you know that at 12:00 pm, the temperature was 12 F and that it decreases by a steady rate of 2 F each hour.

This means that at 1:00 pm, the temperature was 12-2 = 10 F
At 2:00 pm, the temperature was 10-2 = 8 F
At 3:00 pm, it was 8-2 = 6 F
.......and so on

You can follow this sequence to the timing you are looking for.
7 0
3 years ago
So at this rate how many pints can you afford for $20.00
julia-pushkina [17]

Since 1 pint cost 4$

20/4=5

So 5 pints for $20

4 0
3 years ago
You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

6 0
2 years ago
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