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3 years ago

For each box and whisker plot, identify whether the data contains any outliners

Mathematics

1 answer:

Galina-37 [17]3 years ago

7 0

Answer:

ok I can get the money from me and I don't want to be a part of the team and I will be there for sure I will get it to you tomorrow morning to get the kids to the park and ride at work but I can do it

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Which answer choice represents an expanded form of 3 (4+8)?

Sonbull [250]

Answer:

Step-by-step explanation:

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4 years ago

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HELP ASAP!!! make sure to submit a pic of the graph!

gavmur [86]

Answer:

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3 years ago

When the sum of a number plus 3 is squared, it is 9 more than the sum of the number plus 2 when squared.

Komok [63]

$(x+3)^2=(x+2)^2+9\\
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3 0

3 years ago

Alice has $560 in the bank. She uses 1/4 of it to buy a coat and 2/3 of the remaining money to buy a present for her mother. How

Alex

Answer: $140

Step-by-step explanation:

1/4 × 560 to buy a coat

= 0 25 x 560 = $140

Remaining balance after purchase of coat will be $560 - $140 = $420

Next, he uses 2/3 of $420 for present

2/3 × 420

0.667 x 420

= $280 for present

Alice's balance will be remaining balance after purchase of coat - money for present

= $420 - $280

=$140

I hope this is clear, please mark as brainliest answer. Thank you.

5 0

3 years ago

Find an equation of the tangent line to the hyperbola x2 a2 − y2 b2 = 1 at the point (x0, y0).

nalin [4]

The equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

To find the tangent to the hyperbola $\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }$ at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, $\frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }$

So, at (x₀, y₀)

$\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }$

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, $\frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} } \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

So, the equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

Learn more about equation of tangent line here:

brainly.com/question/12561220

3 0

3 years ago