Answer:
1. 1/3 plus 3/6
1/3 is the same as 2/6 so
2/6 plus 3/6 is 5/6
1=5/6
2. 1/4+2/8
2/8 is the same as 1/4 so
1/4 plus 1/4 is 2/4 which is the same as 1/2
2=1/2
Answer:
2 and 12 have same chance of occurring
3 and 11
4 and 10
5 and 9
6 and 8
Each pair has same occurrence chance
Step-by-step explanation:
Here, we are interested in calculating knowing the sums that have an equal chance of occurrence.
Firstly, we need a sample space. A sample space refers to the set of all possible results.
Please check attachment for this
From the sample space we have 36 results;
Let’s list the outcomes and the number of times they all appear;
1- 0
2- 1
3- 2
4-3
5-4
6-5
7- 6
8- 5
9-4
10-3
11- 2
12-1
From the above, we can see that;
2 and 12 have same chance
3 and 11
4 and 10
5 and 9
6 and 8
9514 1404 393
Answer:
a = √(b² +c² -2bc·cos(A))
a = √(c² -b²)
Step-by-step explanation:
For sides a, b, c, and opposite angles A, B, C, the general form of the law of cosines is ...
a² = b² + c² -2bc·cos(A)
An expression for 'a' can be written by taking the square root.
a = √(b² +c² -2bc·cos(A))
__
If you recognize that cos(A) = b/c, then a substitution can be made:
a = √(b² +c² -2(bc)(b/c)) = √(b² +c² -2b²)
a = √(c² -b²) . . . . . . . same as the Pythagorean theorem
Mode- 16 and 5
Median-15
Range- 17
Median-11
Answer:
0.2389
Step-by-step explanation:
The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0.1 mm.
A random sample of 200 wafers is drawn
we are supposed to find What is the probability that the sample mean warpage exceeds 1.305 mm
We will use central limit theorem
According to central limit theorem:
Now we are supposed to find What is the probability that the sample mean warpage exceeds 1.305 mm i.e.P(x>1.305)
Refer the z table
P(z<0.71)=0.7611
Hence the probability that the sample mean warpage exceeds 1.305 mm is 0.2389