85 + 91 + x ÷3 ≥94
You didn't give any choices but this represents the problem.
Answer:
<u>B. The interquartile range (IQR) for town A, 20, is greater than the</u>
<u>IQR for town B, 10.</u>
Step-by-step explanation:
IQR = Q₃ - Q₁
For the town A:
Q₃ = 40 and Q₁ = 20
IQR of town A = 40 - 20 = 20
For the town B:
Q₃ = 30 and Q₁ = 20
IQR of town A = 30 - 20 = 10
Check the given options:
The most statement is appropriate comparison of the spreads is B
<u>B. The interquartile range (IQR) for town A, 20, is greater than the</u>
<u>IQR for town B, 10.</u>
<u />
Simplifying
x + 0.7 = 1 + -0.2x
Reorder the terms:
0.7 + x = 1 + -0.2x
Solving
0.7 + x = 1 + -0.2x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '0.2x' to each side of the equation.
0.7 + x + 0.2x = 1 + -0.2x + 0.2x
Combine like terms: x + 0.2x = 1.2x
0.7 + 1.2x = 1 + -0.2x + 0.2x
Combine like terms: -0.2x + 0.2x = 0.0
0.7 + 1.2x = 1 + 0.0
0.7 + 1.2x = 1
Add '-0.7' to each side of the equation.
0.7 + -0.7 + 1.2x = 1 + -0.7
Combine like terms: 0.7 + -0.7 = 0.0
0.0 + 1.2x = 1 + -0.7
1.2x = 1 + -0.7
Combine like terms: 1 + -0.7 = 0.3
1.2x = 0.3
Divide each side by '1.2'.
x = 0.25
Simplifying
x = 0.25
I only know 1 way.
The first equation is linear:

Divide through by

to get

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for

.
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1xy%5Cright%5D%3D%5Csin%20x)


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The second equation is also linear:

Multiply both sides by

to get

and recall that

, so we can write



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Yet another linear ODE:

Divide through by

, giving


![\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%5Csec%20x%5C%2Cy%5D%3D%5Csec%5E2x)



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In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

then rewrite it as

The integrating factor is a function

such that

which requires that

This is a separable ODE, so solving for

we have



and so on.
Answer: D
Step-by-step explanation: