3y + 7y = 11y - 4y = 7y -6
Answer: 7y - 6
:) I think this may be correct
Answer:
The target population of the survey are travelers vacationing.
Answer:
There is a 33.72% probability that the total weight of the passengers exceeds 4500 pounds.
Step-by-step explanation:
Problems of normally distributed samples can be 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:
There are 22 passengers. Passengers average 190 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. This means that
.
What is the approximate probability that the total weight of the passengers exceeds 4500 pounds?
This probability is 1 subtracted by the pvalue of Z when
. So



has a pvalue of 0.6628.
This means that there is a 1-0.6628 = 0.3372 = 33.72% probability that the total weight of the passengers exceeds 4500 pounds.
Answer:
d = 32.893
Step-by-step explanation:
The distance formula is
d =
now we plug in the x and y
d = 
d = 32.893
The value 12 is a good counterexample. It is divisible by 4 (since 12/4 = 3 is a whole number with no remainders or decimal portions) but 12 is not divisible by 8 (note how 12/8 = 1 remainder 4 = 1.5)
To find other examples, add 8 to 12 to get 20. Then add on 8 more to get 28. This pattern continues on forever.