In your question where ask to find the Standard Normal Distribution of the following:
give probabilities for 0<Z<infinity.
For these ranges, you can read directly, for example,
P(Z<1.96)=0.975.
So for #1, you read directly on the line 1.3 and column 0.03.
For #2, we note that the distribution is symmetrical about Z=0, so
P(Z<-2.33) is the same as P(Z>2.33)
which again is the same as
1-P(Z<2.33) because we know that the area under a probability distribution function adds up to 1.
For the remaining questions, work is similar to #2.
The answer would be 56.10
Answer:
In programming, a recursive function is a function that calls itself. Recursion is used very commonly in programming, although many simple examples (including some shown in this section) are actually not very efficient and can be replaced by iterative methods (loops or vectorized code in MATLAB).
Step-by-step explanation:
Let us name the players A,Dave,Zack,Paul,E and F
For the first position there are two candidades ( Zack / Paul )
For the second position there is only one candidate i.e. Dave
For the third place there will be 4 candidates (out of Zack and Paul - 1 as one of them is already taken for the first position and A, E and F total-4)
For the fourth place there will be 3 candidates ( out of the four available candidates in the 3rd place, one will be taken up for 3rd place )
For the fifth place there will be 2 candidates
Finally, for the last place there will be only one candidate left.
On multiplying the no. of available cadidates, we get 2 * 1 * 4 * 3 * 2 * 1 = 48 i.e. option (A)
Please mention minor spelling mistakes
For the second question:
Let the no of dotted marbles be 'x' and no of striped marbles be 'y'
then the equation will become as follows
(y+6)/x = 3
and
(x+6)/y = (2/3)
On solving the equations, we will get x = 10 and y = 24
Total balls = 10+24+6 = 40 (option E)
Answer 3 will be ) For the first edge, he can choose 3 paths
For the second edge he can choose 2 paths for each path of its first edge's path
For the third , he is bounded to move on the paths created by the first and the second edges hence 1 path for each path created by the first and the second edge together
It will be multiplication of all the possibilities of the paths of the three edges differently.........
i.e. 3 * 2 * 1 = 6