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irinina [24]
3 years ago
10

Suppose a company needs temporary passwords for the trial of a new time management software. Each password will have two letters

followed by two digits. The digits 7,8 , and 9 will not be used. So, there are 26 letters and 7 digits that will be used. Assume that the letters and digits can be repeated. How many passwords can be created using this format
Mathematics
1 answer:
DaniilM [7]3 years ago
5 0

Answer:

33124 passwords can be created using this format

Step-by-step explanation:

The password will have the following format:

L - L - D - D

There are 26 letters and 7 digits, so the number of possible outcomes in each position is:

26 - 26 - 7 - 7

How many passwords can be created using this format?

26*26*7*7 = 33124

33124 passwords can be created using this format

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Answer:

15+6=21

Step-by-step explanation:

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Select the correct answer.
yaroslaw [1]

Answer:

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4 0
3 years ago
Integrated circuits consist of electric channels that are etched onto silicon wafers. A certain proportion of circuits are defec
boyakko [2]

Answer:

z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467  

p_v =P(Z>-2.467)=0.0068  

So the p value obtained was a very low value and using the significance level assumed for example \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of circuits that show evidence of undercutting is significantly less than 0.05.  

Step-by-step explanation:

1) Data given and notation  

n=1000 represent the random sample taken

X=33 represent the number of circuits that show evidence of undercutting

\hat p=\frac{33}{1000}=0.033 estimated proportion of circuits that show evidence of undercutting

p_o=0.05 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that they reduce the rate of undercutting to less than 5%.:  

Null hypothesis:p\geq 0.05  

Alternative hypothesis:p < 0.05  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z>-2.467)=0.0068  

So the p value obtained was a very low value and using the significance level assumed for example \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of circuits that show evidence of undercutting is significantly less than 0.05.  

4 0
3 years ago
Expression 1 6 x 7 – 3^2 x 9 + 4^3 =
matrenka [14]

16x7-3^2x9+4^3

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i hope that helps

8 0
3 years ago
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