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2 years ago

(04.01 LC) 2174 Evaluate (5 points)

Mathematics

1 answer:

Ratling [72]2 years ago

7 0

Answer:

Step-by-step explanation:

542.144638404

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Bill is traveling south and drives 45 miles per hour.

pogonyaev

Answer:

3 hours.

Step-by-step explanation:

Let's say that they both start at a point A. If Bill travels south 45 miles in hour and Landin drives 30 miles north in one hour, after one hour, Bill is going to be 45 miles lower than point A and Landin will be 30 miles above point A, so they will be 45 + 30 = 75 miles apart. After two hours, they will be another 75 miles apart, so they will be a total of 75 + 75 = 150 miles apart. After 3 hours, they will be another 75 miles apart, so they will be a total of 150 + 75 = 225 miles apart.

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3 years ago

A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a samp

fenix001 [56]

Answer:

The lower bound of a 99% C.I for the proportion of defectives = 0.422

Step-by-step explanation:

From the given information:

The point estimate = sample proportion $\hat p$

$\hat p = \dfrac{x}{n}$

$\hat p = \dfrac{55}{100}$

$\hat p$ = 0.55

At Confidence interval of 99%, the level of significance = 1 - 0.99

= 0.01

$Z_{\alpha/2} =Z_{0.01/2} \\ \\ = Z_{0.005} = 2.576$

Then the margin of error $E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1-\hat p)}{n}}$

$E = 2.576 \times \sqrt{\dfrac{0.55(1-0.55)}{100}}$

$E = 2.576 \times \sqrt{\dfrac{0.2475}{100}}$

$E = 2.576 \times0.04975$

E = 0.128156

E ≅ 0.128

At 99% C.I for the population proportion p is: $\hat p - E$

= 0.55 - 0.128

= 0.422

Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422

6 0

2 years ago

Find the measure of angle A

otez555 [7]

Answer:

<A=35

Step-by-step explanation:

70+x+79+x+39=180

2x+188=180

2x=-8

x=-8/2 or -4

<A=x+39

<A=-4+39

<A=35

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3 years ago

Helpppppppppppppooppppppppoopppppppppppppp

MrRissso [65]

I need points soo I can post my question an get help my self...

3 0

3 years ago

Read 2 more answers

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

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

7 0

3 years ago