first, lets distribute the equation. 5(n^2+n)-3n(2n^2+4n-2) would become
5n^2 +5n -6n^3 -12n^2 +6n
Now we add the like terms.
-6n^3 - 7n^2 +11n
Answer:
0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 590 hours.
This means that
Find the probability of a bulb lasting for at most 605 hours.
This is the pvalue of Z when X = 605. So
has a pvalue of 0.8413
0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.
3x² - 5x + 2
= 3x² - 3x - 2x + 2
= 3x(x-1) - 2(x-1)
= (x-1)(3x-2)
Answer:
0.0618.
Step-by-step explanation:
14-18= -4
-4/2.6= -1.538
Use the Z-table to find the proportion that aligns with 1.54 (round up because of the 0.008)
The intersection between 1.5 and 0.04 will be 0.0618.
Raise the unit selling price .25 cents from $2.50 to $2.75