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Sav [38]
1 year ago
7

Part E

Mathematics
1 answer:
Ronch [10]1 year ago
3 0

The option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.

<h3>How to determine the best option to choose?</h3>

The complete question is added as an attachment

From the complete question, the probability in part D is

Probability = 33.5%

The options to choose one from are:

  • Option 1: He can accept the tie, and the game is over.
  • Option 2: He can make one more throw. Jake wins if he earns at least 30 points on the throw, but he loses if he earns less than 30 points.

The calculated probability is less than 0.5.

This means that Jake is more likely to lose than win

So, the best option for Jake is option (A) walk away

Hence, the option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.

Read more about probability at:

brainly.com/question/24756209

#SPJ1

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Answer:

a. \:  \:  \frac{ {4x}^{6} }{ {y}^{3} }  \\

b. \:  \:  \frac{ {8x}^{3} }{ {y}^{ 1\frac{1}{3} } }  \\

Step-by-step explanation:

<h3>a. </h3>

{( {4x}^{ - 2} y)}^{ - 3}  \\  {4x}^{ - 2 \times  - 3}  \:  \: {y}^{1 \times  - 3}  \\  {4x}^{6}  {y}^{ - 3}  \\  \frac{ {4x}^{6} }{ {y}^{3} }  \\

<h3>b.</h3>

{( {8x}^{6} {y}^{ - 3})  }^{ \frac{1}{2} }  \\  {8x}^{6 \times  \frac{1}{2} } \:  \:  {y}^{ - 3 \times  \frac{1}{2} }   \\  {8x}^{3}  {y}^{  - \frac {3}{2} }  \\  \frac{ {8x}^{3} }{ {y}^{ 1\frac{1}{3} } } \\

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A linear equation with an infinite number of solutions is called?
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Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us
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Answer:

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

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".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

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3 years ago
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