Answer:
he would need 7 boxes
Step-by-step explanation:
61/9= 6 with a remainder of 7 therefore instead of 6 boxes you would need 7 to put the rest of the laptops in
Answer:
thank you
Step-by-step explanation:
can I get brainiest please
Answer:
The area lies to the right of the z-score 0.48 means all the values greater than it. This can be calculated on a graphing calculator using the function normCdf, where
- Lower bound = 0.48
- Upper bound = 9999
- Mean = 0
- Standard deviation = 1
<u>The result would be normCdf(0.48,9999,0,1) ≈ </u><u>0.315614</u>
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The area lies to the left of the z-score 0.79 means all the values less than it. This can be calculated on a graphing calculator using the function normCdf, where
- Lower bound = -9999
- Upper bound = 0.79
- Mean = 0
- Standard deviation = 1
<u>The result would be normCdf(-9999,0.79,0,1) ≈</u><u> 0.785236</u>
Answer:
0.087
Step-by-step explanation:
Given that there were 17 customers at 11:07, probability of having 20 customers in the restaurant at 11:12 am could be computed as:
= Probability of having 3 customers in that 5 minute period. For every minute period, the number of customers coming can be modeled as:
X₅ ~ Poisson (20 (5/60))
X₅ ~ Poisson (1.6667)
Formula for computing probabilities for Poisson is as follows:
P (X=ₓ) = ((<em>e</em>^(-λ)) λˣ)/ₓ!
P(X₅= 3) = ((<em>e</em>^(-λ)) λˣ)/ₓ! = (e^-1.6667)((1.6667²)/3!)
P(X₅= 3) = (2.718^(-1.6667))((2.78)/6)
P(X₅= 3) = (2.718^(-1.6667))0.46
P(X₅= 3) = 0.1889×0.46
P(X₅= 3) = 0.086894
P(X₅= 3) = 0.087
Therefore, the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am is 0.087.
