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4 years ago

Jace gathered the data in the table. He found the approximate line of best fit to be y = –0.7x + 2.36.

2 answers:

6 0

Okay but I'm like lowkey positive the answer is -1.14

y = -0.7x + 2.36
x = 5

y = (-0.7 x 5) + 2.36
y = -3.50 + 2.36
y = -1.14

alex41 [277]4 years ago

3 0

Answer with explanation:

The Approximate line of best fit is given by the equation:

y = –0.7 x + 2.36

We have to find , value of ordinate that is y, when ,abscissa is 5.

Putting , x=5 , in the above equation ,to get the value of y

y = -0.7 × 5 + 2.36

= -3.5 +2.36

= -1.14

→None of the option.

if the equation has been like this

y = -0.7 x +0.36

y = -0.7 × 5 + 0.36

= -3.5 +0.36

= -3.14

Option A: -3.14

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2 years ago

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Answer: Josh = 19.8 hours, Danny = 10.8 hours

Step-by-step explanation:

$Josh: \dfrac{1}{x+9}\\\\\\Danny: \dfrac{1}{x}\\\\\\Together: \dfrac{1}{7}\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\\dfrac{1}{x+9}\quad +\qquad \dfrac{1}{x}\qquad = \qquad \dfrac{1}{7}\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(1)(-63)}}{2(1)}\\\\\\.\quad =\dfrac{5\pm\sqrt{25+252}}{2}\\\\\\.\quad =\dfrac{5\pm\sqrt{277}}{2}\\\\\\.\quad =\dfrac{5\pm 16.6}{2}\\\\\\x =\dfrac{5+16.6}{2}\qquad x=\dfrac{5-16.6}{2}\\\\\\x=\dfrac{21.6}{2}\qquad \qquad x=\dfrac{-11.6}{2}\\\\\\x=10.8 \qquad \qquad x=-5.8$

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Danny: x --> 10.8

****************************************************************************************

2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3 (x = -1 is an extraneous solution so is not valid)

2d) No Solution (x = 1 is an extraneous solution so is not valid)

Here is the work for 2a. Follow this format for b, c, & d

$\dfrac{3}{k^2-8x+12}=\dfrac{k}{k-2}-\dfrac{4}{k-6}\\\\\text{Since the denominator cannot equal zero, then } k \neq2\ and\ k\neq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}$

$x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(5)}}{2(1)}\\\\\\.\quad =\dfrac{10\pm\sqrt{100-20}}{2}\\\\\\.\quad =\dfrac{10\pm\sqrt{80}}{2}\\\\\\.\quad =\dfrac{10\pm8.9}{2}\\\\\\x=\dfrac{10+8.9}{2}\qquad x=\dfrac{10-8.9}{2}\\\\\\x=\dfrac{18.9}{2}\qquad x=\dfrac{1.1}{2}\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}$

8 0

3 years ago