The total number of combinations without restrictions is
10 * 10 * 10 * 10 * 10 = 100000
Problem A
10 * 9 * 8 * 7 * 6 = 30240
the number of ways the combination can be set up with no repetition.
The probability of such an event is P = 30240 / 100000 = 0.3024
Problem B
The number of ways that can be done with repetition and no zeros is
9 * 9 * 9 * 9 * 9 = 59049
The probability of such an event is
P = 59049 / 100000 = 0.59049
Answer:
The minimum score of those who received C's is 67.39.
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:

If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?
This is X when Z has a pvalue of 1-0.695 = 0.305. So it is X when Z = -0.51.




The minimum score of those who received C's is 67.39.
Answer:
63720
Step-by-step explanation:
yes
We know that the final price after a discount of 20% is $12,590. If the original price was x, the price after discount is 80%<span> of x (</span>100%<span> - 20%). OP (original price) multiplied by (1-20/100) equals $15,737.5
I hope this helps.</span>
Answer:
The p value for this case is given by:
Since the p value is lower than the significance level we have enough evidence to conclude that the true proportion is significantly higher than 0.5 at 5% of significance.
Step-by-step explanation:
Information given
n=165 represent the random sample selected
55 represent the students indicated a belief that such software unfairly targets students
X =165-55= 110 represent students who NOT belief that such software unfairly targets students
estimated proportion of students who NOT belief that such software unfairly targets students
is the value that we want to check
represent the significance level
z would represent the statistic
represent the p value
System of hypothesis
We want to check if the majority of students at the university do not believe that it unfairly targets them, and the system of hypothesis are:
Null hypothesis:
Alternative hypothesis:
The statistic for this case is given by:
(1)
After replace we got:
The p value for this case is given by:
Since the p value is lower than the significance level we have enough evidence to conclude that the true proportion is significantly higher than 0.5 at 5% of significance.