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zimovet [89]
3 years ago
11

What percentage of the mile times range from 7.25 to 9.375?

Mathematics
1 answer:
Step2247 [10]3 years ago
8 0

1. We assume, that the number 9.375 is 100% - because it's the output value of the task.

2. We assume, that x is the value we are looking for.

3. If 9.375 is 100%, so we can write it down as 9.375=100%.

4. We know, that x is 7.25% of the output value, so we can write it down as x=7.25%.

5. Now we have two simple equations:

1) 9.375=100%

2) x=7.25%

where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:

9.375/x=100%/7.25%

6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for what is 7.25% of 9.375

9.375/x=100/7.25

(9.375/x)*x=(100/7.25)*x       - we multiply both sides of the equation by x

9.375=13.793103448276*x       - we divide both sides of the equation by (13.793103448276) to get x

9.375/13.793103448276=x

0.6796875=x

x=0.6796875

now we have:

7.25% of 9.375=0.6796875

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3 years ago
Express jenny’s age in terms of a if jenny is three years less than half of Anne’s age.
hram777 [196]

Answer:

If Anne’s age is represented by the variable a then express Jenny’s age in terms of a if Jenny is three years less than half of Anne’s age

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2 years ago
HELP ME PLEASE
solmaris [256]

Answer:

The initial mass of the sample was 16 mg.

The mass after 5 weeks will be about 0.0372 mg.

Step-by-step explanation:

We can write an exponential function to model the situation.

Let the initial amount be A. The standard exponential function is given by:

P(t)=A(r)^t

Where r is the rate of growth/decay.

Since the half-life of Palladium-100 is four days, r = 1/2. We will also substitute t/4 for t to to represent one cycle every four days. Therefore:

\displaystyle P(t)=A\Big(\frac{1}{2}\Big)^{t/4}

After 12 days, a sample of Palladium-100 has been reduced to a mass of two milligrams.

Therefore, when x = 12, P(x) = 2. By substitution:

\displaystyle 2=A\Big(\frac{1}{2}\Big)^{12/4}

Solve for A. Simplify:

\displaystyle 2=A\Big(\frac{1}{2}\Big)^3

Simplify:

\displaystyle 2=A\Big(\frac{1}{8}\Big)

Thus, the initial mass of the sample was:

A=16\text{ mg}

5 weeks is equivalent to 35 days. Therefore, we can find P(35):

\displaystyle P(35)=16\Big(\frac{1}{2}\Big)^{35/4}\approx0.0372\text{ mg}

About 0.0372 mg will be left of the original 16 mg sample after 5 weeks.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%287%20-%208%29%20%5Ctimes%202%20%7B%7D%5E%7B2%7D%20" id="TexFormula1" title="(7 - 8) \times 2
Diano4ka-milaya [45]
Answer: -4
Work: 7-8 = -1 x 4
7 0
3 years ago
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