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3 years ago

11

What would be a equation for (5,2) and (0,7)?

1 answer:

Genrish500 [490]3 years ago

6 0

$\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{7}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{7}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{5}}}\implies \cfrac{5}{-5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-1}(x-\stackrel{x_1}{5}) \\\\\\ y-2=-x+5\implies y = -x+7$

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schepotkina [342]

Answer:

150.88

Step-by-step explanation:

just gotta fine the area for each square using the formula base times height with a good old calculator and add them all together, hopefully that's correct

7 0

3 years ago

How do you write the sum in the simplest form 5/12 + 9/12=

Yuliya22 [10]

14/12 because if u go on calculator and do 5 divided by 12 + 9 divided by 12 u get 1.166667 or something and then u do 14/12 and it’s the same number.

3 0

3 years ago

Read 2 more answers

A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when

Genrish500 [490]

Answer:

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

C) CI = (57.53 , 59.07)

D) n = 239.63

Step-by-step explanation:

a)

given data:

mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 25Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025,

Zc = Z(α/2) = 1.96

$ME = Zc * σ \sqrt{n}$

$ME = 1.96 * 3 \sqrt{25}$

ME = 1.18

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{25} , 58.3 + 1.96 * 3\sqrt{25})$

CI = (57.12 , 59.48)

b)

Given data:

mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

$ME = zc * σ \sqrt{n}$

$ME = 1.96 * 3\sqrt{100}$

ME = 0.59

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{100} , 58.3 + 1.96 * 3\sqrt{100})$

CI = (57.71 , 58.89)

c)

sample mean, $\bar X = 58.3$

sample standard deviation, σ = 3

sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

$ME = Zc * σ \sqrt{n}$

$ME = 2.58 * 3\sqrt{100}$

ME = 0.77

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 2.58 * 3\sqrt{100} , 58.3 + 2.58 * 3/\sqrt{100}$

CI = (57.53 , 59.07)

D)

Given data:

Significance Level, α = 0.01,

Margin or Error, E = 0.5,

σ = 3

The critical value for α = 0.01 is 2.58.

for calculating population mean we used

$n \geq (zc *σ/E)^2$

$n = (2.58 * 3/0.5)^2$

n = 239.63

7 0

3 years ago

In 1 hr, a plane can travel 440 mi into the wind and 480 mi with the wind. Find the wind speed and the rate of the plane in stil

nexus9112 [7]

Answer:

the plane would fall and crash at 150mph

Step-by-step explanation:

8 0

3 years ago

The answers is what I got but it said it was wrong. can anyone hhelp find the right answer?

klio [65]

Answer:

but its right

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers