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expeople1 [14]
3 years ago
15

Whats next 2 numbers in this pattern 4,20,100,500.....

Mathematics
1 answer:
Korolek [52]3 years ago
7 0
The next two patterns are 2,500 12,500
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3 out of five picks are orange . If 12 picks are orange , how many picks are there in all?
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20 picks in total .

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Shawn says that 3.22 is a rational number. Leah says 3.22 is a repeating decimal. Who is correct and why?
MAVERICK [17]

Answer:

Hi, There! Leah Is Correct

Step-by-step explanation:

A fraction is a repeating decimal if the prime factors of the denominator of the fraction in its lowest form do not only contain 2s and/or 5s or do not have any prime factors at all.

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2 years ago
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Find the product of (3x + 7y)(3x − 7y)
nalin [4]

Answer:

9x^2 - 49y^2

Step-by-step explanation:

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2 years ago
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A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
3 years ago
in 4 minutes Liaa can type 140 words. If she has an essay taht is 700 words long , about how long will it take Liana to type it
Zanzabum

Answer: Lol your in collage I don't believe it anyway the answer is 20 min on the dot  

6 0
2 years ago
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