The probability that a randomly selected x-value from the distribution will be in the interval:
- P(35 < x < 45) = 0.6827 and,
- P(30 < x < 40) = 0.47725
<h3>What is the probability of a normal distribution?</h3>
The probability of a normal distribution can be determined from the symmetrical curve between 1 to 100%.
From the information given:
- Mean = 40
- Standard deviation = 5
To determine the probability that a randomly selected x-value is in the given interval:
![\mathbf{P(35 < x < 45) = P[Z\le 1] -P[Z\le -1]}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%2835%20%3C%20x%20%3C%2045%29%20%3D%20P%5BZ%5Cle%201%5D%20-P%5BZ%5Cle%20-1%5D%7D)
Using normal distribution table:
P(35 < x < 45) = 0.8414 - 0.1587
P(35 < x < 45) = 0.6827
![\mathbf{P(30 < x < 40) = P[Z\le0]-P[Z\le -2]}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%2830%20%3C%20x%20%3C%2040%29%20%3D%20P%5BZ%5Cle0%5D-P%5BZ%5Cle%20-2%5D%7D)
Using normal distribution table:
P(30 < x < 40) = 0.5 - 0.02275
P(30 < x < 40) = 0.47725
Learn more about the probability of a normal distribution here:
brainly.com/question/4079902
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Answer:
Step-by-step explanation:
Rewrite:
![\dfrac{1}{\sqrt[3]{n^2}}=9\\\\9\sqrt[3]{n^2}=1](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Csqrt%5B3%5D%7Bn%5E2%7D%7D%3D9%5C%5C%5C%5C9%5Csqrt%5B3%5D%7Bn%5E2%7D%3D1)
Cube both sides:
Divide both sides by 729:
Take the square root of both sides:
Hope this helps!
Answer:
9/20
Step-by-step explanation:
so since trentons pet eats 4/5 of the food each day and marias eats 1.25 each day so we cant put 4/5 and 1.25 to see wat the awnser is so we make 1.25 a fraction 1 1/4 but the demoninators arent the same so what we do is make them the same so i multiply 4x5 and get 20 that is out new denominator and 4x4 is 16 so the 4/5 is now 16/20 and the 1 1/4 is 1 5/20 then subtract but since we have a whole number we subtract that first but they dont go together but we do it anyway and get 4 left over we put the 4 into the 5/20 and get 9/20 and thers ur awnser
