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oee [108]
3 years ago
13

(Please show your work) (1 kwh = 3400 BTU’s 1 gallon = 8 pounds of water) The water coming into a dishwasher from a water heater

is 50 degrees F. It has to be heated to 100 degrees. The dishwasher uses 10 gallons of water and .5 kwh of electrical energy to operate. How much total energy in BTU’s is needed to heat the water and run the dishwasher?
Mathematics
1 answer:
shtirl [24]3 years ago
8 0

Answer:

5700 BTU

Step-by-step explanation:

Data provided in the question:

1 kwh = 3400 BTU’s

1 gallon = 8 pounds of water

Initial temperature = 50°F

Final temperature = 100°F

Water used by dishwasher = 10 gallons

Electrical energy used = 0.5 kwh

Now,

Water used in pounds = 10 × 8 = 80 lb

Total electrical energy used in BTU = 0.5 × 3400 BTU     [ as 1 kwh = 3400 BTU]

Total electrical energy used in BTU = 1700 BTU

Energy used by water = Water used by dishwasher × Change in temperature × Specific heat of water

[  Specific heat of water = 1 BTU / °F.lb ]

Thus,

Energy used by water = 80 × 1 × ( 100 - 50 )

= 4000 BTU

Hence,

Total energy used = 1700 BTU + 4000 BTU = 5700 BTU

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Extra Info:
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The y intercept is (0,6). Think backwards in terms of the pattern going on. 
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2 years ago
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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.
zmey [24]

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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Answer:

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Over [174]

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Answer:    \bold{\dfrac{35}{187}}

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