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3 years ago

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Question 3 A school wants to know what type of new food to add to the cafeteria menu for next year that most students will like.

They ask every 4th person at a football game, what food they would like. Will this sample lead to valid inferences? Explain why or why not.

1 answer:

Alik [6]3 years ago

8 0

Yes because if you are asking every 4th kid then you are have a chance of one of of four

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X-9= -6<br> Please help me answer this question ASAP

Ganezh [65]

Answer:

its 3

Step-by-step explanation:

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3 years ago

Solve the equation: 3.2/(z-1/2) = 2.667/(z+1/3)

Luden [163]

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Z = 2489/533

Step-by-step explanation:

I went on photo math and solved it

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3 years ago

Holly has watched 39% of a movie that is 122 minutes long.

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48 minutes

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39% of 122 is ...

0.39 × 122 = 47.58 ≈ 48

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3 years ago

NEED HELP ASAP FAST

liubo4ka [24]

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the probability is 50% or 1/2 (equally likely and equally unlikely)

That is because it does not state the amount of socks inside the draw; meaning the answer has to be theoretical.

If the probability is the same, than it has to equal to 100

so we can conclude that 1/2 of the drawer is blue socks and 1/2 is white socks.

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3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago