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kvasek [131]
2 years ago
13

22. The mean of 15 observations is 32. Find the resulting mean, if each observation is: (i) increased by 3 (ii) decreased by(v)

increased by 60 percent
​
Mathematics
1 answer:
Lubov Fominskaja [6]2 years ago
8 0

Answer:

35 ;

32 + v

51.2

Step-by-step explanation:

If each observations in a particular set is increased or decreased by a uniform or fixed number, the mean of the resulting output will also increase by the same amount

Hence,

For 15 observations whose mean is 32

If each observation is increased by 3, then the resulting mean will also increase by 3 ; hence, new mean = 32 + 3 = 35

If each observation is decreased by v ; the resulting mean would also decrease by v ;

32 - v

If each observation is increased by 60% ; the resulting mean will be ;

(1 + 60%) * initial mean

(1 + 0.6) * 32

1.6 * 32 = 51.2

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Answer:

108 : 145

Step-by-step explanation:

253 - 108 = 145

Ratio of those who receive 2 weeks of paid vacation is 108

Ratio of those paid vacation is not 2 weeks is 145

108 : 145

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3 years ago
What is the answer explain please
eduard

Answer:

Pattern B

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2 years ago
How many 12 hours periods (half days) will pass after 25 hours? Enter only the whole number of period that have passed?
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Step-by-step explanation:

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A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of
Eva8 [605]

Answer:

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=250000 represent the sample mean  

s=37500 represent the standard deviation for the sample

n=5 sample size  

\mu_o =210250 represent the value that we want to test  

\alpha=0.1 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)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:  

Null hypothesis:\mu \leq 210250  

Alternative hypothesis:\mu > 210250  

Compute the test statistic  

We don't know the population deviation, so for this case 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".  

We can replace in formula (1) the info given like this:  

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

Conclusion

Since is a one right tailed test the p value would be:  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

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jok3333 [9.3K]

Answer:

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Step-by-step explanation:

If ur able to use a calculator there is a slope calculator online and it's easy to use

8 0
2 years ago
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