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2 years ago

9

Jordan goes to a high school with 1900 students. He surveyed a random sample of 240 students about their favorite sport, and 150

said that football was their favorite sport. Jordan reported to the school newspaper that football is the favorite sport of 1188 students. What inference did Jordan make? PLSS HELP QUICK !!

1 answer:

just olya [345]2 years ago

7 0

Answer: answer b

Step-by-step explanation:

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What is 5(-x) equal to?

Gnoma [55]

Answer:

?

Step-by-step explanation:

where are the options?

4 0

2 years ago

Find the equation of the line.<br> Use exact numbers.

luda_lava [24]

Answer:

y = -2x + 5

Step-by-step explanation:

Looking at the line, every time it moves one unit to the right, it goes down two units. Therefore, the slope is -2. Since the y-intercept is 5, the equation of this line is y = -2x+5.

I hope this helped!

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4 0

1 year ago

Why is the answer to this integral's denominator have 1+pi^2

ss7ja [257]

It comes from integrating by parts twice. Let

$I = \displaystyle \int e^n \sin(\pi n) \, dn$

Recall the IBP formula,

$\displaystyle \int u \, dv = uv - \int v \, du$

Let

$u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Then

$\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn$

Apply IBP once more, with

$u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Notice that the ∫ v du term contains the original integral, so that

$\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)$

$\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I$

$\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n$

$\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C$

6 0

2 years ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago

How many terms are there in a geometric series if the first term is 3, the common ratio is 4, and the sum of the series is 1,023

tatiyna

In Geometric Progression, we know S(n) = a.(rⁿ - 1 ) / r - 1
Here, a = 3, r = 4 a(n) = 1,023
1023 = 3 (4ⁿ - 1) / (4 - 1)
3(4ⁿ - 1) = 1023 * 3
4ⁿ - 1 = 3069 / 3
4ⁿ = 1023 + 1
4ⁿ = 1024
4⁵ = 1024

In short, Your Answer would be 5 Terms

Hope this helps!

6 0

3 years ago