Answer:
The only one that looks familiar is B sorry >-<
Step-by-step explanation:
Answer:
I think the answer is the third one
I believe you meant:
If the average, mode, and range of 5 different numbers are all 8, find the sum of all possible values and the second largest number
Answer:
Sum of all possible values = 40
Second largest number= 8
Explanation:
Since our mean, mode, and range is 8, our sum is 40 and our second largest number is 8.
The sum of all the numbers is 40 since our mean is 8 and there are 5 numbers(8×5)
To find possible values, we already know the mode is 8 and so we know 8 must appear more than once. If we have 8 twice then we are left with 24(sum is 40). We have 3 numbers left and the range is 8 so we must have a maximum number and minimum number that has a difference of 8 and a third number, all adding up to 24(sum left). Within 24, only 12-4+8 satisfies the condition, hence our numbers are: 4,8,8,8,12.
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 25235
For the alternative hypothesis,
µ > 25235
This is a right tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 100,
Degrees of freedom, df = n - 1 = 100 - 1 = 99
t = (x - µ)/(s/√n)
Where
x = sample mean = 27524
µ = population mean = 25235
s = samples standard deviation = 6000
t = (27524 - 25235)/(6000/√100) = 3.815
We would determine the p value using the t test calculator. It becomes
p = 0.000119
Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.
