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alexandr1967 [171]

3 years ago

5

What is the difference between the weight of 5.595.59 lb and the mean of the weights? b. How many standard deviations is that

[the difference found in part (a)]? c. Convert the weight of 5.595.59 lb to a z score. d. If we consider weights that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the weight of 5.595.59 lb significant?

1 answer:

icang [17]3 years ago

6 0

They weight different

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago

BartSMP [9]

Answer:

B.

Step-by-step explanation:

Causation is often confused with correlation because they are so similar. B would mean that pollution is caused from something (lets say greenhouse gases for example) which then results in more cases of heart disease.

Causation is when the first variable may bring the second into existence or may cause the incidence of the second variable to fluctuate. So, causation is the capacity of one variable to influence another. Which, would make B, a suitable candidate.

8 0

3 years ago

How do i do this?????????

bagirrra123 [75]

(x+3)-2 that is your answer I know because I'm great at this unit

8 0

3 years ago

Hi... can you tell me is this formula is right or wrong??

Llana [10]

Answer:

this formula is right

6 0

3 years ago

Based on the given angle measures, which triangle has side length measures that could be correct?

Likurg_2 [28]

Answer

The triangle in the figure 4 is correct .

Reason

by using the trignometric identity

$tan\theta = \frac{Perpendicular}{Base}$

Thus

$tan 30^{\circ} = \frac{Perpendicular}{Base}$

$tan 30 ^{\circ} =\frac{1}{\sqrt{3}}$

Now in the figure (1)

$tan 30^{\circ} = \frac{13.9}{8}$

$\frac{1}{\sqrt{3}} = \frac{13.9}{8}$

on simplify

$0.5774 \neq 1.7375$

thus side length measures in the figure (1) is not correct .

Now in the figure (2)

$tan30^{\circ} =\frac{16}{8} \\ \frac{1}{\sqrt{3}} =\frac{16}{8} \\ 0.577 \neq 2$

thus side length measures in the figure (2) is not correct .

Now in the figure (3)

$tan 30 ^{\circ} = \frac{8}{16} \\ \frac{1}{\sqrt{3}} =\frac{8}{16} \\ 0.577\neq0.5$

thus side length measures in the figure (3) is not correct .

Now in the figure (4)

$tan30^{\circ} = \frac{8}{13.9} \\ \frac{1}{\sqrt{3}} =\frac{8}{13.9}\\ 0.577 = 0.576(approx)$

Therefore the figure (4) is correct triangle has side length measures that could be correct

Hence proved

8 0

3 years ago

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