Pl help is for homwork
1 answer:
Answer:
Sabrina bought 3 pounds of bananas.
Step-by-step explanation:
Given equations are:
a + b = 8 Eqn 1
0.95a + 1.10b = 8.05 Eqn 2
As we have to find the number of pounds of bananas bought, we will eliminate the variable a from the equation.
Multiplying Eqn 1 by 0.95
0.95(a+b=8)
0.95a+0.95b=7.60 Eqn 3
Subtracting Eqn 3 from Eqn 2
(0.95a+1.10b)-(0.95a+0.95b)=8.05-7.60
0.95a+1.10b-0.95a-0.95b=0.45
0.15b = 0.45
Dividing both sides by 0.15
Hence,
Sabrina bought 3 pounds of bananas.
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Answer:
Answer:
There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.
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
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so .
What is the probability that a fisherman catches a spotted seatrout within the legal limits?
They must be between 10 and 30 inches.
So, this is the pvalue of the Z score of subtracted by the pvalue of the Z score of
X = 30
has a pvalue of 0.9772
X = 10
has a pvalue of 0.00135.
This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.