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3 years ago

What single percentage change is equivalent to a 12% decrease followed by a 13% decrease?

Mathematics

1 answer:

drek231 [11]3 years ago

6 0

Answer:

The % decrease will be 24.44%

Step-by-step explanation:

Let's say that the original number is x.

(u/x)*100=12

u=12x/100

x-(12x/100)=(100x-12x)/100=88x/100

Now (u/88x/100) *100=13

(50u/44x)*100=13

u=13*44x/5000

u=572x/5000

(88x/100)-(572x/5000)=(4400x-572x)/5000=3828x/5000=0.7656x

Thus x-0.7656x=0.2444x

Therefore there will be a decrease by 24.44%

You can evaluate this equation by using real time values, which will be a much easier calculation.

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One half of a number added to two thirds of the number is 21. Find the number and show your work

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The number would be 18.
Half of 18 = 9
Two thirds of 18 = 12
9 + 12 = 21

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Q. During one particular sales day, 75% of the store's customers

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60-75%= 45
So 60-45=15
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I ate 7/8 of birthday cake.how much of cake has not been eaten?

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3 years ago

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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Question 10(Multiple Choice Worth 1 points)

Mila [183]

Answer:
—6

Explanation:
In an expression with a negative number and a positive number, the negative will come on top. This meaning the product (end piece) will be negative. In this problem despite the signs we know 3•2 is 6. Now since one number is negative (3) and the other is positive (2) negative will come out on top.

* KEEP IN MIND *
+ and + = positive
- and + = negative

Easy Memo for Multiplication and Divison:
“Same sign add, different sign subtract”
Add = Positive (+)
Subtract = Negative (-)

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