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denis23 [38]
3 years ago
12

Val’s whole house central air conditioner uses 4,000 watts when running Val runs the ac 6 hrs per summer day electricity cost (1

5cents) (1 kilowatt-hr) how much does Val ac cost to run for a summer month of 30 days
Mathematics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

Val ac cost 108000 dollar  for the summer month of 30 days

Step-by-step explanation:

Given

Electricity consumed in a day by running AC is 4000 * 6 = 24000 watts-hr

Cost of electricity is 15 cents/kilowatt-hr

Cost of electricity consumed in a month

24000*30*0.15= 108000 dollars

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Answer:

a = (-7)

Step-by-step explanation:

Firstly clear the bracket...,

-6(-2 + a) = 12 - 6a

Then substitute the simplified version in place of bracket...,

; 12 - 6a = 54

; -6a = 54 - 12

; -6a = 42...then divide both sides by (-6)

Therefore...., a = (-7)

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Answer:

A.

Step-by-step explanation:

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A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.
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Answer:

t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18    

df=n-1=40-1=39  

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

p_v =P(t_{(39)}>2.18)=0.0177  

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, and then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation  

\bar X=2.2 represent the sample mean

s=0.29 represent the sample standard deviation

n=40 sample size  

\mu_o =2.1 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 2,1, the system of hypothesis would be:  

Null hypothesis:\mu \leq 2.1  

Alternative hypothesis:\mu > 2.1  

If we analyze the size for the sample is > 30 but 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".  

Calculate the statistic

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

t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18    

P-value

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

df=n-1=40-1=39  

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

p_v =P(t_{(39)}>2.18)=0.0177  

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, and then the population mes seems to be higher than 2.1 at 5% of significance

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