Answer:
Part a.)
C = 3N
Note: Any variables could be used in this equation. It is important
that the students be able to define them.
Part b.)
3c + 750 = 1200 (or equivalent representation)
‐750 ‐750
3c = 450
c = 150
750 + 3c ≥ 1200
‐750 ‐750
3c ≥ 450
c ≥ 150
Parts c. and d.)
3(200) + 750 = T
600 + 750 = T
1350 = T
Step-by-step explanation:
plssss give brainliest and thanks
Answer:
<h2>94 hours</h2>
Step-by-step explanation:
<h3>3 days = 3×24</h3><h3> =72</h3><h3>total hours= 72 + 22</h3><h3> = 94</h3>
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.
Hello,
Please, see the attached files.
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8x+13 is the simplified version