On the lines below write a word problem that involves division and would have a correct answer of 4/7

1 answer:

6 0

Jim was driving down a country road, late in the afternoon. He noticed that the farm he was passing had a peculiar way of separating it's sheep, male on one side and female on the other. There were 70 sheep in all, and 30 males. What percent of the sheep population were female?

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Which one is bigger 3.5km or 3,000 km

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Converting kilometers to meters could never be simpler with our accurate length converter! ... Meters and kilometers both belong to the metric length scale, with these two units of measurement being on the larger side. ... 1 Kilometer equals 1000 Meters (1km = 1000m) ... 3 Kilometers equal 3000 Meters (3km = 3000m) ...

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Which function is the inverse of function 1?<br> f(x) = 9.12 - 12

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Solving quadratic equation using the completing the square (a).<br>9k²+5k=2

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The standard deviation of a point estimator is the _____. a. standard error b. sample statistic c. point estimate d. sampling er

ivann1987 [24]

The standard deviation of a point estimator is the <u>standard error.</u>

"The point estimate is used to estimate the exact value of any parameter. For example, an internet poll says that 45% of Americans are concerned for the environment."

"Point estimates is a good measure but the problem with this parameter is that the sample of the statistics may vary so that there is a chance of sampling error."

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To learn more about standard error here

brainly.com/question/13179711

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1 year ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

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2 years ago