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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 ![\sigma = 15, \mu = 590](https://tex.z-dn.net/?f=%5Csigma%20%3D%2015%2C%20%5Cmu%20%3D%20590)
Find the probability of a bulb lasting for at most 605 hours.
This is the pvalue of Z when X = 605. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{605 - 590}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B605%20-%20590%7D%7B15%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
has a pvalue of 0.8413
0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.
Answer:
16
Step-by-step explanation:
Comment if you want explanation
Cos 157.5º=-cos (180º-157.5º)=-cos 22.5=-cos(45/2)
cos Ф/2=⁺₋√((1+cosФ)/2).
In this case 157.5º is in the second quadrant, therefore we use the following equation:
cos Ф/2=-√((1+cosФ)/2). (we will have a negative number)
cos 157.5º=-cos (45/2)=-√((1+cos 45º)/2)
=-√((1+√2/2)/2)
=-√((2+√2)/4)
=-√(2+√2) / 2 (≈-0.92387...)
Answer: cos 157.5º= -√(2+√2) / 2
Answer:
(A) -8
Step-by-step explanation:
∆=b^2 -4ac= (-4)^2-4(2)(3)= -8
Since discriminant<0, then the equation has no roots.