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Mice21 [21]
3 years ago
15

A ball with mass m kg is thrown upward with initial velocity 28 m/s from the roof of a building 17 m high. Neglect air resistanc

e Use g = 9.8 m/s. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. Fend Click If you would like to Show Work for this question: Open Show Work LINK TO TEXT
Mathematics
1 answer:
Nutka1998 [239]3 years ago
3 0

a. At its maximum height, the ball has zero vertical velocity, so

-\left(28\dfrac{\rm m}{\rm s}\right)^2=2(-g)(y_{\rm max}-17\,\mathrm m)

\implies y_{\rm max}=57\,\mathrm m

b. The ball's height in the air y at time t is given according to

y=17\,\mathrm m+\left(28\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

Solve for t when y=0:

17+28t-4.9t^2=0\implies t\approx6.3\,\mathrm s

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Answer:

D. Congruent after a dilation

Step-by-step explanation:

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4 years ago
The simple interest on an investment of $8000 for one year is $360.
elena55 [62]
I = PRT
$360 = $8000 x R X 1
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4 years ago
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2
nikklg [1K]

Answer:

n=97

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma=0.01)

We know that the margin of error for a confidence interval is given by:

Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=.05 and \alpha/2=0.025

Using the normal standard table, excel or a calculator we see that:

z_{\alpha/2}=1.96

If we solve for n from formula (1) we got:

\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}

n=(\frac{z_{\alpha/2} \sigma}{Me})^2

And we have everything to replace into the formula:

n=(\frac{1.96(0.05)}{0.01})^2 =96.04

And if we round up the answer we see that the value of n to ensure the margin of error required \pm=0.01 Liters is n=97.

5 0
4 years ago
Exercise 6
Lady_Fox [76]

a)\\\\x^2-2x-4=0\\\\x^2-2x=4\\\\x^2-2x+1^2=4+1^2\\\\(x-1)^2=5\\\\x-1=\pm\sqrt5\\\\\underline{x=\pm\sqrt5+1\approx-1.24\ or\ 3.24}\\\\\\b)\\\\x^2-4x-2=0\\\\x^2-4x=2\\\\x^2-4x+2^2=2+2^2\\\\(x-2)^2=6\\\\x-2=\pm\sqrt6\\\\\underline{x=\pm\sqrt6+2\approx-0.45\ or\ 4.45}

c)\\\\x^2+8x+5=0\\\\x^2+8x=-5\\\\x^2+8x+4^2=-5+4^2\\\\(x+4)^2=11\\\\ x+4=\pm\sqrt{11}\\\\\underline{x=\pm\sqrt{11}-4\approx-7.32\ or\ -0.68}\\\\\\x)\\\\x^2+22x-15=0\\\\x^2+22x=15\\\\x^2+22x+11^2=15+11^2\\\\(x+11)^2=136\\\\x+11=\pm\sqrt{136}\\\\\underline{x=\pm\sqrt{136}-11\approx-22.66\ or\ 0.66}

e)\\\\x^2-7x+10=0\\\\x^2-7x=-10\\\\x^2-7x+(\frac72)^2=-10+(\frac72)^2\\\\x^2-7x+(\frac72)^2=-10+(\frac72)^2\\\\(x-\frac72)^2=-\frac{20}4+\frac{49}4\\\\ (x-\frac72)^2=\frac{29}4\\\\x-\frac72=\pm\sqrt{\frac{29}4}\\\\ \underline{x=\pm\frac{\sqrt{29}}2+\frac72\approx6.19\ or\ 0.81}\\\\\\f)\\\\x^2+5x-2=0\\\\x^2+5x=2\\\\x^2+5x+(\frac52)^2=2+(\frac52)^2\\\\(x+\frac52)^2=2+6.25 \\\\x+\frac52=\pm\sqrt{8.25}\\\\\underline{x=\pm\sqrt{8.25}-\frac52\approx-5.37\ or\ 0.37}

g)\\\\x^2+9x-4=0\\\\x^2+9x=4\\\\x^2+9x+(\frac92)^2=4+(\frac92)^2\\\\(x+\frac92)^2=4+20.25\\\\x+\frac92=\pm\sqrt{24.25}\\\\\underline{x=\pm\sqrt{24.25}-4.5\approx-9.42\ or\ 0.52}

h)\\\\x^2-3x-6=0\\\\x^2-3x=6\\\\x^2-3x+(\frac32)^2=6+(\frac32)^2\\\\(x-\frac32)&^2=6+2.25\\\\x-\frac32=\pm\sqrt{8.25}\\\\ \underline{x=\pm\sqrt{8.25}+1.5\approx4.37\ or-1.37}

3 0
3 years ago
Many residents of suburban neighborhoods own more than one car but consider one of their cars to be the main family vehicle. The
horrorfan [7]

Answer:

95.4% of family vehicles is between 1 and 3 years old.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2

Standard Deviation, σ = 6 months = 0.5 year

We are given that the distribution of age of cars is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(family vehicles is between 1 and 3 years old)

P(1 \leq x \leq 3)\\\\ = P(\displaystyle\frac{1 - 2}{0.5} \leq z \leq \displaystyle\frac{3-2}{0.5}) = P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.977 -0.023 = 0.954= 95.4\%

P(1 \leq x \leq 3) = 95.4%

95.4% of family vehicles is between 1 and 3 years old.

3 0
3 years ago
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