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BartSMP [9]
2 years ago
7

2.5y=5 what is the answer to the equation?

Mathematics
2 answers:
postnew [5]2 years ago
7 0

Answer:2

Step-by-step explanation:

2.5(2)

=5

DedPeter [7]2 years ago
5 0

Answer:

y=.5

Step-by-step explanation:

Set 2.5 as 25 and weal come back to the deci. 25y=5 is the same as 25/5 there for you get the answer 5 but now you add the decimal getting you to .5 . welcome

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An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample
kolbaska11 [484]

Answer:

a) 654.16-2.01\frac{164.55}{\sqrt{52}}=608.29    

654.16+2.01\frac{164.55}{\sqrt{52}}=700.03    

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=52-1=51

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that t_{\alpha/2}=2.01

Replacing we got:

654.16-2.01\frac{164.55}{\sqrt{52}}=608.29    

654.16+2.01\frac{164.55}{\sqrt{52}}=700.03    

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, and we use an estimator of the population variance the value of 175 replacing into formula (b) we got:

n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189

6 0
3 years ago
The mean of a set of five test scores is m. what is the sum of the five test scores?
FinnZ [79.3K]
Mean = Sum of data / Number of data

Mean of 5 test scores = m

Sum of 5 test scores / 5 = m

Sum of 5 test scores = 5m

Hence, the answer is 5m.
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3 years ago
What is the average distance of earth to the sun in km
dybincka [34]
The average distance is 3.23!! I found this question very interesting thanks!
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2 years ago
Last question plz help
Viktor [21]
the answer is 1,5 OK
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