Answer:
Probabilty= 7.66*10^-3
Step-by-step explanation:
A box contains 65 balls numbered from 1 to 65.
9 balls are drawn with replacement.
Probabilty of selecting a number = 1/65
Probabilty of selecting a number
= 0.0154
Probabilty of at least 2 numbers out of 9 selected to have Same number
Probabilty= 9C2(0.0154)^2(0.9846)^7
Probabilty= 7.66*10^-3
Answer:
The Domain is that X is between -6 and 6
The range is that y is between -8 and 4
Step-by-step explanation:
Look at the two endpoints of the function. You can see that the left-most point is at (-6, 4) and the right-most point is at (6, -8)
I am very stuck on this answer simplify 8/12
Answer:
Weights of at least 340.1 are in the highest 20%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

a. Highest 20 percent
At least X
100-20 = 80
So X is the 80th percentile, which is X when Z has a pvalue of 0.8. So X when Z = 0.842.




Weights of at least 340.1 are in the highest 20%.
Answer:
y-determinant = 2
Step-by-step explanation:
Given the following system of equation:
Let's represent it using a matrix:
![\left[\begin{array}{ccc}1&2\\1&-3\end{array}\right] = \left[\begin{array}{ccc}5\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C1%26-3%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
The y‐numerator determinant is formed by taking the constant terms from the system and placing them in the y‐coefficient positions and retaining the x‐coefficients. Then:
![\left[\begin{array}{ccc}1&5\\1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%265%5C%5C1%267%5Cend%7Barray%7D%5Cright%5D%20)
y-determinant = (1)(7) - (5)(1) = 2.
Therefore, the y-determinant = 2