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2 years ago

Please Simplify (3 x-10) ^3

Mathematics

1 answer:

masya89 [10]2 years ago

8 0

Answer: 27 x 3 − 270 x 2 + 900 x − 1000

Step-by-step explanation: Expand the expression.

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A school band has 64 members. The band marches in the shape of a rectangle.

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Step-by-step explanation:

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3 years ago

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If angle 1 and angle 2 are adjacent angles, then do angle 1 and angle 2 form a linear pair?

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Yes, if two angles are adjacent, that makes them a linear pair.

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2 years ago

A large on-demand video streaming company is designing a large-scale survey to determine the mean amount of time corporate execu

Diano4ka-milaya [45]

Answer:

$n=(\frac{2.262(3)}{0.25})^2 =736.79 \approx 737$

So the answer for this case would be n=737 rounded up to the nearest integer

Step-by-step explanation:

We have the following info given:

$\bar X = 12$ the sample mean

$s = 3$ the sample deviation

$n = 10$ represent the sample size

The degrees of freedom are given by:

$df =n-1 =10-1=9$

The margin of error is given by this formula:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{t_{\alpha/2} s}{ME})^2$ (5)

The confidence level is 95% and the critical value for this case would be given by $t_{\alpha/2}=2.262$ , replacing into formula (5) we got:

$n=(\frac{2.262(3)}{0.25})^2 =736.79 \approx 737$

So the answer for this case would be n=737 rounded up to the nearest integer

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2 years ago

Explorers on a planet with a thin atmosphere want to measure the acceleration of gravity at its surface. So they use a spring gu

Murljashka [212]

Answer:

The value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$ .

Step-by-step explanation:

We know that the position function is given by

$s(t)=30t-\frac{g_s}{2} t^2$

Velocity is defined as the rate of change of position. Therefore,

$v(t)=\frac{d}{dt}(s(t))$

So, the velocity function is

$v(t)=\frac{d}{dt}(s(t))\\\\v(t)=\frac{d}{dt}(30t-\frac{g_s}{2} t^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(30t\right)-\frac{d}{dt}\left(\frac{g_s}{2}t^2\right)\\\\v(t)=30-g_st$

When a body reaches a vertical velocity of zero, this is the maximum height of the body and then gravity will take over and accelerate the object downward. Thus,

$v(t)=30-g_st=0\\\\g_s=\frac{30}{t}$

We know that the ball bearing reaches its maximum height 21 seconds after being launched (t = 21 s).

So, the value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$ .

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