He spends 32 hours in the gym in an 8 week period. (8*4)
Answer:
Ok so the first one is right and the second one is right for the first question
For the second Question your also right
Step-by-step explanation:
Question 3
12/7 is greater than 6/13 and it's asking what numbers would be incorrect meaning less than 6/13 (unmark 12/7)
1/10 is correct since 6/13 is greater than 3 / 65 (It's what you get when you multiple 1/10 with 6/13)
Unmark 4/3
Unmark 3
and 13/13 is just equal to 6/13 (leave unmarked)
Question 4
I don't know kiddo it's been a while
<u></u>
For the answer to the question above asking an<span> office supplies salesperson sold $1,284 in week 1, $976 in week 2, $489 in week 3 and $1,042 in week 4.
Is this an incomplete question or you are asking for the total sales?
should I add them up? The answer would be 3,791</span>
Answer:
1963.2 pounds (lbs.)
Step-by-step explanation:
Things to understand before solving:
- - <u>Normal Probability Distribution</u>
- The z-score formula can be used to solve normal distribution problems. In a set with mean ц and standard deviation б, the z-score of a measure X is given by:

The Z-score reflects how far the measure deviates from the mean. After determining the Z-score, we examine the z-score table to determine the p-value associated with this z-score. This p-value represents the likelihood that the measure's value is less than X, or the percentile of X. Subtracting 1 from the p-value yields the likelihood that the measure's value is larger than X.
- - <u>Central Limit Theorem</u>
- The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean ц and standard deviation б , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean ц and standard deviation

As long as n is more than 30, the Central Limit Theorem may be applied to a skewed variable. A specific kind of steel cable has an average breaking strength of 2000 pounds, with a standard variation of 100 pounds.
This means, ц = 2000 and б = 100.
A random sample of 20 cables is chosen and tested.
This means that n = 20, 
Determine the sample mean that will exclude the top 95 percent of all size 20 samples drawn from the population.
This is the 100-95th percentile, or X when Z has a p-value of 0.05, or X when Z = -1.645. So 
- By the Central Limit Theorem


<h3>Answer:</h3>
The sample mean that will cut off the top 95% of all size 20 samples obtained from the population is 1963.2 pounds.