Answer:
HHH, W = 3-0 = 3
HHT, W= 2-1=1
HTH, W= 2-1=1
THH, W=2-1 =1
HTT, W= 1-2=-1
THT, W= 1-2=-1
TTH, W=1-2=-1
TTT, W=0-3 = -3
So then the sample space for W is:
![S= [-3,-1,1,3]](https://tex.z-dn.net/?f=%20S%3D%20%5B-3%2C-1%2C1%2C3%5D)
Just 4 possible values from 8 possible combinations for the 3 random tosses
Step-by-step explanation:
For this case we define W as the random variable who represent the number of heads minus the number of tails in three tosses of a coin.
W= # heads- # coins
Since we toss a coin 3 times we have 2*2*2= 8 possible results. We can list the results and the corresponding values for W like this:
HHH, W = 3-0 = 3
HHT, W= 2-1=1
HTH, W= 2-1=1
THH, W=2-1 =1
HTT, W= 1-2=-1
THT, W= 1-2=-1
TTH, W=1-2=-1
TTT, W=0-3 = -3
So then the sample space for W is:
![S= [-3,-1,1,3]](https://tex.z-dn.net/?f=%20S%3D%20%5B-3%2C-1%2C1%2C3%5D)
Just 4 possible values from 8 possible combinations for the 3 random tosses
The answer is C or 12 because it has to equal to 180 so 5(12) + 5 = 65 and 65 + 115 = 180
Answer:
16
Step-by-step explanation:
The number would be:
= 2 + (-6 + 20)
= 2 - 6 + 20 = 16
Hope this right because your question is not very clear! :3
If a line is drawn across the shape connecting two of the angles it makes two triangles. The sum of the angles in a triangle is 180 degrees.
Since this shape makes two triangles the sum of the inside angles is 2 * 180 = 360 degrees
(2x - 17) + 90 + x + 59 = 360
combine like terms
3x + 132 = 360
3x = 360 - 132
3x = 228
x = 76
Step-by-step explanation:
Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.
Type ni xi si
Graded 42 0.486 0.187
No-fines 42 0.359 0.158
To find - a. Formulate the above in terms of a hypothesis testing problem.
b. Give the test statistic and its reference distribution (under the null hypothesis).
c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.
Proof -
a.)
Hypothesis testing problem :
H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
b)
Test statistic :




⇒Z(cal) = 3.3687
Z(tab) = 1.96
As Z (cal) > Z(tab)
So, we reject H0 at 5% Level of significance
p-value = 0.99962
Hence
There is significant difference in mean conductivity at the two materials.