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sergiy2304 [10]
3 years ago
9

The percent of working students increased 8.1 to 40.5 what was the present prior to increase

Mathematics
1 answer:
Dmitrij [34]3 years ago
6 0

Answer:

  32.4

Step-by-step explanation:

prior + 8.1 = 40.5 . . . . . . seems to model the problem statement

prior = 32.4 . . . . . . . subtract 8.1 from both sides

Prior to the increase the percent was 32.4.

_____

<em>Comment on the problem statement</em>

When you're talking about a percentage increase in a percentage, it is almost never clear whether you're talking about the percentage of the underlying number, or the percentage of the percentage.

Here, we assume the 8.1 is a percentage of working students, not a percentage of the percentage of workings students. If you actually intend the latter, the percentage before the increase was about 37.465%.

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bob charges 6.25 per hour for washing cars, If he works 8 1/2 hours a day, about how much money would bob make in a 5 day week?
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Answer:

265.65

Step-by-step explanation:

6.25*8 =50

6.25/2=3.13

50+3.13=53.13

53.13*5=265.65

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2. A particular resource class must perform at a level consistent with the rest of the school to avoid being discontinued. On th
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Answer:

t=\frac{71.86-78}{\frac{17.52}{\sqrt{14}}}=-1.311  

p_v =P(t_{13}  

If we compare the p value and a significance level assumed for example \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the scores is NOT significant less than 78.  

Step-by-step explanation:

Data given and notation  

\bar X=71.86 represent the sample mean  

s=17.52 represent the standard deviation for the sample  

n=14 sample size  

\mu_o =78 represent the value that we want to test  

\alpha 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.  

We need to conduct a hypothesis in order to determine if the mean is less than 78, the system of hypothesis would be:  

Null hypothesis:\mu \geq 78  

Alternative hypothesis:\mu < 78  

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".  

Calculate the statistic  

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

t=\frac{71.86-78}{\frac{17.52}{\sqrt{14}}}=-1.311  

Calculate the P-value  

First we need to find the degrees of freedom given by:

df=n-1=14-1=13

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

p_v =P(t_{13}  

Conclusion  

If we compare the p value and a significance level assumed for example \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the scores is NOT significant less than 78.  

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3 years ago
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