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3 years ago

8

To find out whether the students really like the school's lunch, the school randomly selects half of the student body to survey

about the school's lunch.
Is this sample of students likely to be biased or unbiased?

biased

unbiased

1 answer:

AleksAgata [21]3 years ago

3 0

I think it will be unbiased

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Find x. Round your answer to the nearest tenth of a degree.

Y_Kistochka [10]

Answer:

x ≈ 42.5

Step-by-step explanation:

We use tan∅ to solve this:

tanx° = 11/12

x° = $tan^{-1}(11/12)$

x = 42.5104°

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3 years ago

The price of a gallon of gasoline increased from 2.50 a gallon to 2.75 a gallon what was the percent increase

Vera_Pavlovna [14]

The increase in price was 10%
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2.5+.25=2.75
hope this helps

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2 years ago

Evaluate the spherical coordinate integral

expeople1 [14]

Rewrite the equations of the given boundary lines:

<em>y</em> = -<em>x</em> + 1 ==> <em>x</em> + <em>y</em> = 1

<em>y</em> = -<em>x</em> + 4 ==> <em>x</em> + <em>y</em> = 4

<em>y</em> = 2<em>x</em> + 2 ==> -2<em>x</em> + <em>y</em> = 2

<em>y</em> = 2<em>x</em> + 5 ==> -2<em>x</em> + <em>y</em> = 5

This tells us the parallelogram in the <em>x</em>-<em>y</em> plane corresponds to the rectangle in the <em>u</em>-<em>v</em> plane with 1 ≤ <em>u</em> ≤ 4 and 2 ≤ <em>v</em> ≤ 5.

Compute the Jacobian determinant for this change of coordinates:

$J=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}1&1\\-2&1\end{bmatrix}\implies|\det J|=3$

Rewrite the integrand:

$-3x+4y=-3\cdot\dfrac{u-v}3+4\cdot\dfrac{2u+v}3=\dfrac{5u+7v}3$

The integral is then

$\displaystyle\iint_R(-3x+4y)\,\mathrm dx\,\mathrm dy=3\iint_{R'}\frac{5u+7v}3\,\mathrm du\,\mathrm dv=\int_2^5\int_1^45u+7v\,\mathrm du\,\mathrm dv=\boxed{333}$

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3 years ago

Consider the equation 2x+8y=12

rusak2 [61]

For a you have to solve 2 solutions , first 2x=12
6=x then you plug that back in, 12+8y=12
8y=0
y=0
(6,0)

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3 years ago

The equation of the graphed line in point-slope form is _______, and its equation in slope-intercept form is .

Ivenika [448]

Answer:

Point-slope form equation line

$(y-3)=(-0.6)\times (x+2)$

Slope-intercept form is:

$y=-0.6x+1.8$

Step-by-step explanation:

Point-slope form equation line passing through points $(x_1,y_1)(x_2,y_2)$

Slope = m = $\frac{y-2-y_1}{x_2-x_1}$

slope-intercept form is:

y = mx +c

where : c = intercept of on axis

$(y-y_1)=m\times (x-x_1)$

Equation of line passing through (-2,3) and (3,0).

$m=\frac{0-3}{3-(-2)}=-0.6$

$(y-3)=-0.6\times (x-(-2))$

Point-slope form equation line

$(y-3)=(-0.6)\times (x+2)$

For slope-intercept form open the brackets of Point-slope form equation l:

$y-3=-0.6x-1.2$

So, the slope-intercept form is:

$y=-0.6x+1.8$

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3 years ago