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Fudgin [204]
3 years ago
7

While attending a college football game, you think you can guess the attendance within 10% of the actual value. If the announced

attendance is 65,000, within what number of people would your guess have to be in order for you to support your claim?
Mathematics
1 answer:
Softa [21]3 years ago
3 0
10% of 65,000 is 6,500. So the range will be between 65,000+6,500 (71,500) and 65,000-6,500 (58,500)

58,500-71,500
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How to solve 4x-10>x/3-12 step by step showing to to get rid of fractions
zhenek [66]

Answer:

x > -6/11

Step-by-step explanation:

4x - 10 > x/3 - 12

~Simplify

4x - 10 > 1/3x - 12

~Add 10 to both sides

4x > 1/3x - 2

~Subtract 1/3x to both sides

11/3x > -2

~Multiply 3/11 to both sides

x > -6/11

Best of Luck!

4 0
3 years ago
Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help
ki77a [65]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867785

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}


Make a trigonometric substitution:

\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}


so the integral (i) becomes

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}


Now, substitute back for t = arcsin(x²), and you finally get the result:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}          ✔

________


You could also make

x² = cos t

and you would get this expression for the integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}

\mathsf{=\dfrac{\pi}{4}}         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)

6 0
3 years ago
Guys please help me i really need this right answer
Ipatiy [6.2K]

Answer:

Step 2: Commutative Property

Step 3: Associative Property

Step-by-step explanation:

According to the Commutative Property of Addition: Two real numbers can be added in either order.

a + b = b + a

Therefore, she used the Commutative Property in Step 2.

In Step 3, she used the Associative Property, in which the grouping of the addends does not change the sum.

4 0
3 years ago
Word problem for 24- ( 6+3)
Diano4ka-milaya [45]
24-(6+3) =
24-9=
15 is your answer. Just simply break the math problem down. It’ll help
7 0
4 years ago
Read 2 more answers
Polynomials <br>change of subject formulas​
SOVA2 [1]
Example:
To make ‘u’ the subject of the formula in

V = u + at

*subtract ‘at’ to both sides*
V -at = u + at - at

Result:
V - at = u

So the answer would be

u = V - at
8 0
4 years ago
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