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Gennadij [26K]
3 years ago
11

A tennis player gets an ace on 35% of his serves. Out of 80 serves about how many aces will he get

Mathematics
2 answers:
kari74 [83]3 years ago
7 0

Answer:

28

Step-by-step explanation:

His aces on 80 serve = 35% of 80

= 35/100 × 80

= 28

Harrizon [31]3 years ago
4 0

Answer:

He will get 28 aces out of 80 serves if he get aces on 35% of them.

Step-by-step explanation:

First do

35 x 80 = 2,800

then

2,800 / 100 = 28

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Use the figure below to determine the relationship between the lengths of MR and MS.
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A computer randomly selects a number from the given set. {2, 4, 6, 9, 16, 44, 51, 60, 78} What is the probability that a number
nekit [7.7K]
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3 years ago
Please answer the 1 question in the picture, thanks
Anestetic [448]

Answer:

40

Step-by-step explanation:

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2 years ago
Read 2 more answers
Please see the image and help with this homework problem. thank you!
dimulka [17.4K]

we have the function

E=\sin \frac{\pi}{14}t

Part a

For t=7

substitute in the given function

\begin{gathered} E=\sin \frac{\pi}{14}7 \\ E=1 \end{gathered}

For t=14

\begin{gathered} E=\sin \frac{\pi}{14}14 \\ E=0 \end{gathered}

For t=21

\begin{gathered} E=\sin \frac{\pi}{14}21 \\ E=-1 \end{gathered}

For t=28

\begin{gathered} E=\sin \frac{\pi}{14}28 \\ E=0 \end{gathered}

For t=35

\begin{gathered} E=\sin \frac{\pi}{14}35 \\ E=1 \end{gathered}

Observation: The values of E varies from -1 to 1, including the zero

Part B

Remember that

The Period goes from one peak to the next

so

Period=2pi/B

B=pi/14

Period=(2pi)/(pi/14)=2pi*14/pi=28

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6 0
1 year ago
It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken to estimate
Hitman42 [59]

Answer:

The minimum sample size is n = 75 so that the desired margin of error is 5 or less.                          

Step-by-step explanation:

We are given the following in the question:

Population variance = 484

Standard deviation =

\sigma^2 = 484\\\sigma =\sqrt{484} = 22

Confidence level = 0.95

Significance level = 0.05

Margin of error = 5

Formula:

Margin of error =

E = z\times \dfrac{\sigma}{\sqrt{n}}\\\\n = (\dfrac{z\times \sigma}{E})^2

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

Putting values, we get

E\leq 5\\\\1.96\times \dfrac{22}{\sqrt{n}} \leq 5\\\\\sqrt{n} \geq 1.96\times \dfrac{22}{5}\\\\n \geq (1.96\times \dfrac{22}{5})^2\\\\n \geq 74.373

Thus, the minimum sample size is n = 75 so that the desired margin of error is 5 or less.

3 0
2 years ago
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