Answer:
b = 
Step-by-step explanation:
b +
= 1
← change to an improper fraction
b +
= 
Multiply through by 12 ( the LCM of 3 and 4 ) to clear the fractions
12b + 8 = 15 ( subtract 8 from both sides )
12b = 7 ( divide both sides by 12 )
b = 
Answer:
Probability = 0.12025
Step-by-step explanation:
P (Am) = 1/50 = 0.02 {Magazine ad}
P (At) = 1/8 = 0.125 {Television ad}
P (Am ∩ At ) = 1/100 = 0.01 {Both ads}
P (Am U At) = P (Am) + P (At) - P (Am ∩ At )
= 0.02 + 0.125 - 0.01
P (Am U At) = 0.135 {Person sees either ad}
P (Am' ∩ At') = 1 - P (Am U At)
P (Am' ∩ At') = 1 - 0.135 = 0.865 {Person sees none ad}
Prob (Purchase) = Prob (Purchase with ad) + Prob (purchase without ad)
P (P/ A) = 1/4 = 0.25 , P (P / A') = 1/10 = 0.1
P (P) = (0.25) (0.135) + (0.1) (0.865)
= 0.03375 + 0.0865
0.12025
Answer: a constant
Step-by-step explanation:
A constant is a numerical expression like 2, 0,78 etc .
A variable is a alphabetical expression that vary like a,b,c,d,x,y,z.
An expression can consists of both numerical and alphabets and also any arithmetic expression like x, 2abc, 6a+2c etc
A term consist of either numbers and variables multiplied together or only numbers or variable like 2xy, x, 2ab etc.
X is all (a term, a variable , an expression) except a constant because a constant is a numerical expression.
To solve this formula for T, divide both sides of the original equation by PR:
I PRT
------ = --------- => T = I / (PR)
PR PR
Please note: Because the formula I = PRT involves neither addition nor subtraction, the final formula for T cannot involve either addition nor subtraction. That leaves:
T = I P/R
T= IPR
The second formula here is incorrect; we cannot solve I = PRT for T simply by rearranging the order of the variables. This leaves T = I P/R as a possible answer, but this answer does not agree with my T = I / (PR). Please double check to ensure that you have copied down the four possible answers correctly.
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.