Answer:
There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.
Step-by-step explanation:
Problems dealing with normally distributed samples can be addressed using the z-score formula.
For a set with the mean
and a standard deviation
, the z-score for a measure X is calculated by

Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.
For this case
Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus 
What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?
That translates to 
Thus



has a p-value of 0.7549.
<pthis indicates="" that="" src="https://tex.z-dn.net/?f=P%28X%20%5Cleq%2050%29%20%3D%200.7549" id="TexFormula10" title="P(X \leq 50) = 0.7549" alt="P(X \leq 50) = 0.7549" align="absmiddle" class="latex-formula">.
Additionally, we have that


There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.
</pthis>
Hey! Hi! Hola! Aloha!
When rounding 1199.28856995 to the nearest tenth, the result is 119.3.
I hope this is helpful!
Response:
The answer is 1/13
To explain step-by-step:
We are notinformed about the first (01) twelve (12) cards that are shown; hence, the probability that the thirteenth (13) card dealt is a King is equivalent to the probability of the first card being a King, or specifically any designated card dealt being a King, equating to:
=4/52
=
1/13 FINAL RESULT
Among the options provided, the statement that holds true is the third one: "Nina makes 20x dollars each week at the clothing store." The information given explicitly mentions that x denotes the total hours she has worked within a week. This assertion renders the other statements incorrect.