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1 year ago

9

Find the variance of: 6 6 7 8 10 10

1 answer:

marta [7]1 year ago

7 0

<h3>3.368</h3>

Step-by-step explanation:

I hope this helps!!!!!

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I attached a pic please help

umka2103 [35]

Function P: y intercept of 5 (0,5), no slope
Function Q: y=- \frac{1}{3} x+4, m=- \frac{1}{3}

A: Both functions do NOT have the same slope.
B: Both functions do NOT have negative slope.
C: Functions will NOT have same input when y=0
D. The functions will have different outputs when x=0 (Function P: (0, 5) and Function Q: (0, 4)

6 0

3 years ago

A company manufactured a 16-oz box of cereal. Boxes are randomly weighted to ensure the correct amount. If the discrepancy in we

Katena32 [7]

15.85 oz to 16.15 since .15 ounces less than 16 is 15.85 and .15 ounces more than 16.15

3 0

3 years ago

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Please help me with this question.

Vesna [10]

Answer:

3 to 9

second is 2.5 to 15

Step-by-step explanation:

3 0

2 years ago

Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e

svet-max [94.6K]

ANSWER

$y = 2x -4$

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

$x = 5 + ln(t)$

$y = {t}^{2} + 5$

From the first equation we make t the subject to get;

$x - 5 = ln(t)$

$t = {e}^{x - 5}$

We put it into the second equation.

$y = { ({e}^{x - 5}) }^{2} + 5$

$y = { ({e}^{2(x - 5)}) } + 5$

We differentiate to get;

$\frac{dy}{dx} = 2 {e}^{2(x - 5)}$

At x=5,

$\frac{dy}{dx} = 2 {e}^{2(5 - 5)}$

$\frac{dy}{dx} = 2 {e}^{0} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

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

$y - 6 = 2(x - 5)$

$y = 2x - 10 + 6$

$y = 2x -4$

Without eliminating the parameter,

$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

$\frac{dy}{dx} = \frac{ 2t}{ \frac{1}{t} }$

$\frac{dy}{dx} = 2 {t}^{2}$

At x=5,

$5 = 5 + ln(t)$

$ln(t) = 0$

$t = {e}^{0} = 1$

This implies that,

$\frac{dy}{dx} = 2 {(1)}^{2} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

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

$y - 6 = 2(x - 5) =$

$y = 2x -4$

5 0

3 years ago

Find the equation of a line that goes through the point (1,-1) and is parallel to the line y = -6x + 1

Katena32 [7]

y=-6x+5

the slope needs to stay -6 so the line stays parallel, with the slope going down 6, right 1, the y-int would have to be 5

5 0

2 years ago

Read 2 more answers