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Hatshy [7]
3 years ago
9

Selina and three of her classmates each wrote out the steps for finding an approximate percent change. They all agreed on the fi

rst and fourth step:
1. Find the positive difference between the original number and the new number.
2. ________________________
3. ________________________
4. Round the answer if necessary.

They did not agree on steps 2 and 3. Which instructions are correct for steps 2 and 3?
Mathematics
2 answers:
Naddik [55]3 years ago
5 0

Answer: 2. Divide the difference by the original number.

3. Multiply it by 100.

Step-by-step explanation:

Since when we find the percentage change in a number then we follow the following steps,

1. find the difference between the new number and the original number.

2. divide the difference by the original number.

3. Multiply it by 100.

4.Round the answer if necessary.

We can understand it with help of the below example.

If a number 25 changes to 75.

Then the percentage change in 25 = \frac{75-25}{25} \times 100=\frac{50}{25}\times 100 = 200%

kotegsom [21]3 years ago
3 0
Step 2 is divide the difference between the original number and the new number by the original number. you will get a decimal.

step 3 convert what you got from step 2 to a percent by multiplying it by 100 and adding %
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Find the inflection point of the function of f(x) = x³ -6x² +12x
Wewaii [24]

Points of inflections are zeroes of the second derivative. We have:

f(x)=x^3-6x^2+12x\implies f'(x)=3x^2-12x\implies f''(x)=6x-12

So, the second derivative equals zero if and only if

f''(x)=0\iff 6x-12=0 \iff 6x=12 \iff x=2

So, this is the only point of inflection of this function.

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2 years ago
Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming poo
kolezko [41]

Let <em>x</em> and <em>y</em> be the unit rates at which one large pump and one small pump works, respectively.

Two large/one small operate at a unit rate of

(1 pool)/(4 hours) = 0.25 pool/hour

so that

2<em>x</em> + <em>y</em> = 0.25

One large/three small operate at the same rate,

(1 pool)/(4 hours) = 0.25 pool/hour

<em>x</em> + 3<em>y</em> = 0.25

Solve for <em>x</em> and <em>y</em>. We have

<em>y</em> = 0.25 - 2<em>x</em>   ==>   <em>x</em> + 3 (0.25 - 2<em>x</em>) = 0.25

==>   <em>x</em> + 0.75 - 6<em>x</em> = 0.25

==>   5<em>x</em> = 0.5

==>   <em>x</em> = 0.1

==>   <em>y</em> = 0.25 - 2 (0.1) = 0.25 - 0.2 = 0.05

In other words, one large pump alone can fill a 1/10 of a pool in one hour, while one small pump alone can fill 1/20 of a pool in one hour.

Now, if you have four each of the large and small pumps, they will work at a rate of

4<em>x</em> + 4<em>y</em> = 4 (0.1) + 4 (0.05) = 0.6

meaning they can fill 3/5 of a pool in one hour. If it takes time <em>t</em> to fill one pool, we have

(3/5 pool/hour) (<em>t</em> hours) = 1 pool

==>   <em>t</em> = (1 pool) / (3/5 pool/hour) = 5/3 hours

So it would take 5/3 hours, or 100 minutes, for this arrangement of pumps to fill one pool.

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