Answer:
The proportions differ from those reported in the survey.
Step-by-step explanation:
The Chi-square goodness of fit test would be used to determine whether the proportions differ from those reported in the survey.
The hypothesis for the test can be defined as follows:
<em>H</em>₀: The proportions does not differ from those reported in the survey.
<em>Hₐ</em>: The proportions differ from those reported in the survey.
Assume that the significance level of the test is, α = 0.01.
The Chi-square test statistic is given by:

Consider the Excel sheet provided.
The Chi-square test statistic value is 191.32.
The <em>p</em>-value of the test is:

The <em>p</em>-value of the test is very small. The null hypothesis will be rejected at 1% level of significance.
Thus, concluding that the proportions differ from those reported in the survey.
Step-by-step explanation:
2 : 3 = 6 : x ,
Ratio can be written as
2*3 = 6, we multiply top and bottom of fraction by 3
So value of x=9
4 : 7 = x : 42 ,
7*6 = 42, we multiply top and bottom of fraction by 6
So value of x=24
2x : 48 = 3 : 12,
\frac{3}{12}[/tex] =
2x= 12, so x= 6
12 : 15 = x : 20
,
we multiply top and bottom by 4/3

So value of x=16
Michael will end up on the 30th floor.
Answer:
Step-by-step explanation:
8x + 12 = 3x + 5x + 12
8x + 12 = 8x + 12
Bringing like terms on one side
8x - 8x = 12 - 12
x = 0
Answer: The mean and standard deviation are 567.2 and 89.88 resp.
Step-by-step explanation:
Since we have given that
For 370 parts per million = 7% = 0.07
For 440 parts per million = 10% = 0.10
For 550 parts per million = 49% = 0.49
For 670 parts per million = 34% = 0.34
So, Mean of the carbon dioxide atmosphere for these trees would be
![E[x]=370\times 0.07+440\times 0.1+550\times 0.49+670\times 0.34=567.2](https://tex.z-dn.net/?f=E%5Bx%5D%3D370%5Ctimes%200.07%2B440%5Ctimes%200.1%2B550%5Ctimes%200.49%2B670%5Ctimes%200.34%3D567.2)
And
![E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 0.34=329794](https://tex.z-dn.net/?f=E%5Bx%5E2%5D%3D370%5E2%5Ctimes%200.07%2B440%5E2%5Ctimes%200.1%2B550%5E2%5Ctimes%200.49%2B670%5E2%5Ctimes%200.34%3D329794)
So, Variance would be
![Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16](https://tex.z-dn.net/?f=Var%5C%20x%3DE%5Bx%5E2%5D-E%5Bx%5D%5E2%3D329794-567.2%5E2%3D8078.16)
So, the standard deviation would be

Hence, the mean and standard deviation are 567.2 and 89.88 resp.