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Misha Larkins [42]
3 years ago
5

Please help me I need it

Mathematics
2 answers:
love history [14]3 years ago
7 0
I am pretty sure it’s 120%
levacccp [35]3 years ago
5 0

Answer:

112%

Step-by-step explanation:

percentage change value formula

c2 =  <u>X2  -  X 1</u>       X   <u>100</u>

             X1                   1

WHERE,

C2 = RELATIVE VALUE

X2  = FINAL VALUE  ( TODAY'S VALUE)  = 32,860

X1 = INITIAL VALUE  (1992 VALUE) = 15,500

PERCENTAGE CHANGE = 100

C2 =   <u>32860  -   15500</u>   X   <u>100</u>

                 15500                    1

=    <u>17, 360 </u>       X 100

     15,500                 = 1.12 X 100 = 112%.

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Step-by-step explanation:

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2 years ago
Write an equation that represents a discount of 18% on a retail price of $55. Let p represent the new price.
trapecia [35]

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6 0
3 years ago
Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T w
Alinara [238K]

Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100×(\frac{1}{2} )^{3.322}.

Step-by-step explanation:

The equation for half life is of the form

A = A₀×(\frac{1}{2} )^{\frac{t}{h} }.........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ \frac{t}{h}  and and the equation (2) can be written as

Percentage remaining after 2 half lives is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 }

However if the half life of Technetium-99m is 6 hours then we have for one day

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 *2}

Therefore an expression that represents the number of days until only 10% remains is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{\frac{d}{h}  } = 10 %

(\frac{1}{2} )^{\frac{d}{h}  } =\frac{1}{10}

= ㏑(\frac{1}{2} )^{\frac{d}{h}  }  = ㏑(\frac{1}{10})

= \frac{d}{h}×㏑(\frac{1}{2} ) = ㏑(\frac{1}{10})

\frac{d}{h} = \frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )} = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100×(\frac{1}{2} )^{3.322}

3 0
4 years ago
Which of the following are true about the graph of f(x)=-4x^2?
MrMuchimi

Answer:

the statements that are true are a & d

6 0
3 years ago
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