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Romashka-Z-Leto [24]
3 years ago
8

The Australian Open is the first of the four Grand Slam professional tennis events held each year. Victoria Azarenka beat Maria

Sharapova to win the 2012 Australian Open women’s title. During the tournament Ms. Azarenka serve speed reached kilometers per hour. A list of the Women’s Singles serve speed for the 2012 Australian Open is provided below. Player Serve Speed (km/h) S. Williams 192 S. Lisicki 192 M. Keys 191 L. Hradecka 188 J. Gajdosova 187 J. Hampton 187 B. Mattek-Sands 185 F. Schiavone 185 P. Ormaechea 185 P. Parmentier 185 N. Petrova 183 G. Arn 183 V. Azarenka 182 A. Ivanovic 182 P. Kvitova 179 M. Krajicek 179 V. Dushevina 178 S. Stosur 176 S. Cirstea 176 M. Barthel 175
a. Compute the mean, variance, and standard deviation for the serve speeds.

b. A similar sample of the 20 Women’s Singles serve speed leaders for the 2011 Wimbledon tournament showed a sample mean serve speed of 182.5 kilometers per hour. The variance and standard deviation were 33.3 and 5.77, respectively. Discuss any difference between the serve speeds in the Australian Open and the Wimbledon women’s tournaments.
Mathematics
1 answer:
klasskru [66]3 years ago
5 0

Answer:

asnwer is B

Step-by-step explanation:

mark BRAINLIEST

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If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

3 0
3 years ago
Answer??????????????????????
ch4aika [34]
Here is the answer you are looking for: (7,13/3)

Please vote my answer brainliest. thanks!
5 0
3 years ago
If a change dispenser contains 32 coins consisting of
swat32
<h3>Answer:  19 dimes</h3>

=================================================

Work Shown:

d = number of dimes

q = 32-d = number of quarters

$5.15 = 515 cents

10d+25q = 515

10d+25(32-d) = 515

10d+800-25d = 515

-15d+800 = 515

-15d = 515-800

-15d = -285

d = -285/(-15)

d = 19

There are 19 dimes. You can stop here if you want.

q = 32-d = 32-19 = 13

There are 13 quarters

-------------------------------

Check:

1 dime = 10 cents

19 dimes = 19*10 = 190 cents

1 quarter = 25 cents

13 quarters = 13*25 = 325 cents

total value = 190 cents + 325 cents = 515 cents = $5.15

The answer is confirmed.

6 0
3 years ago
Leslie drank 4/5 pint of juice and Hank drank 7/10 pint of juice. Who drank more
amid [387]
Leslie did. If you multiply 4/5 by two, you get 8/10. 8/10 is greater than 7/10. 
5 0
3 years ago
Read 2 more answers
Question is in the pictures.
Kipish [7]
<h3>Answer:</h3>

x/tan(x) is an even function

sec(x)/x is an odd function

<h3>Explanation:</h3>

<em>x/tan(x)</em>

For f(x) = x/tan(x), consider f(-x).

... f(-x) = -x/tan(-x)

Now, we know that tan(x) is an odd function, so tan(-x) = -tan(x). Using this, we have ...

... f(-x) = -x/(-tan(x)) = x/tan(x) = f(x)

The relation f(-x) = f(x) is characteristic of an even function, one that is symmetrical about the y-axis.

_____

<em>sec(x)/x</em>

For g(x) = sec(x)/x, consider g(-x).

... g(-x) = sec(-x)/(-x)

Now, we know that sec(x) is an even function, so sec(-x) = sec(x). Using this, we have ...

... g(-x) = sec(x)/(-x) = -sec(x)/x = -g(x)

The relation g(-x) = -g(x) is characeristic of an odd function, one that is symmetrical about the origin.

6 0
3 years ago
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