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Tresset [83]
1 year ago
10

. Write a function wordscramble that will receive a word in a string as an input argument. It will then randomly scramble the le

tters and return the result. Here is an example of calling the function: >> wordscramble('fantastic') ans = safntcait
Computers and Technology
1 answer:
inysia [295]1 year ago
4 0

Answer: You can use the function randperm() to generate a random permutation of the indexes. so then you would use the random permutation to reorder the string array

Explanation:

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Find a 95% confidence interval for the mean failure pressure for this type of roof panel.The article "Wind-Uplift Capacity of Re
aliya0001 [1]

Answer:

The 95% confidence interval would be given by (2.351;3.525)    

Explanation:

1) Previous concepts

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".

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".

\bar X represent the sample mean  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=2.938

The sample deviation calculated s=0.472

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=5-1=4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that t_{\alpha/2}=2.78

Now we have everything in order to replace into formula (1):

2.938-2.78\frac{0.472}{\sqrt{5}}=2.351    

2.938+2.78\frac{0.472}{\sqrt{5}}=3.525    

So on this case the 95% confidence interval would be given by (2.351;3.525)    

5 0
2 years ago
Write a script to check command arguments (3 arguments maximum). Display the argument one by one. If there is no argument provid
serious [3.7K]

Answer:-args (

if:args=true

-cont

if:args=false

-cont investigating

if:args=irrelevance

-loop restate args

)

compile exec

Explanation:

7 0
1 year ago
What kind of problems could you run into if you format a cell with the wrong format for the data type?
exis [7]
The answer to the following question

<span>What kind of problems could you run into if you format a cell with the wrong format for the data type?

is:

there is a great possibility that your file format won't open because it has the wrong format</span>
8 0
2 years ago
Read 2 more answers
What is a text feature that could add visual interest and clarity to a procedural document?
rjkz [21]

Bullet points is another

3 0
1 year ago
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) Consider a router that interconnects four subnets: Subnet 1, Subnet 2, Subnet 3 and Subnet 4. Suppose all of the interfaces in
erik [133]

Answer:

Check the explanation

Explanation:

223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.

The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000

Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)

So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26

Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)

and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25

Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)

So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28

4 0
2 years ago
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