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:
90 mi/h
Step-by-step explanation:
Given,
For first 30 miles, her speed is 30 miles per hour,
Let x be her speed in miles per hour for another 30 miles,
Since, here the distance are equal in each interval,
So, the average speed of the entire journey
![=\frac{\text{Average speed for first 30 miles + Average speed for another 30 miles}}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctext%7BAverage%20speed%20for%20first%2030%20miles%20%2B%20Average%20speed%20for%20another%2030%20miles%7D%7D%7B2%7D)
![=\frac{30+x}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B30%2Bx%7D%7B2%7D)
According to the question,
![\frac{30+x}{2}=60](https://tex.z-dn.net/?f=%5Cfrac%7B30%2Bx%7D%7B2%7D%3D60)
![30+x=120](https://tex.z-dn.net/?f=30%2Bx%3D120)
![\implies x = 90](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%2090)
Hence, she needs to go 90 miles per hour for remaining 30 miles.
Answer:
30
Step-by-step explanation:
Two or zero expresses the possible number of positive real solutions for the given polynomial equation.
Answer: Option D
<u>Step-by-step explanation:</u>
Given equation:
![x^{3}-4 x^{2}-7 x+28=0](https://tex.z-dn.net/?f=x%5E%7B3%7D-4%20x%5E%7B2%7D-7%20x%2B28%3D0)
First, we put hit and trial method to find out the one solution. So, if we put x=4 then the above expression will become zero. We can also write the above expression as
![(x-4) \times\left(x^{2}-7\right)=0](https://tex.z-dn.net/?f=%28x-4%29%20%5Ctimes%5Cleft%28x%5E%7B2%7D-7%5Cright%29%3D0)
We know the formula,
, make use of this, we get
![(x-4) \times(x-\sqrt{7}) \times(x+\sqrt{7})=0](https://tex.z-dn.net/?f=%28x-4%29%20%5Ctimes%28x-%5Csqrt%7B7%7D%29%20%5Ctimes%28x%2B%5Csqrt%7B7%7D%29%3D0)
So, ![x=4, \sqrt{7},-\sqrt{7}](https://tex.z-dn.net/?f=x%3D4%2C%20%5Csqrt%7B7%7D%2C-%5Csqrt%7B7%7D)
Hence, from the above expression, we have three values of x as x= 4, 2.64 and -2.64
The answer is: " -5/3 " ; or, write as: " -1 ⅔ " .
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Explanation:
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(-5/6) + (-5/6) = (-5/6) <span>− (5/6) ;
------> {since: "adding a negative" is the same a "subtracting a positive"} ;
------> </span> (-5/6) − (5/6) = (-5 − 5) / 6 ;
= -10/6 = (-10/2) / (6/2) ;
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= " -5/3 " ; or, write as: " -1 ⅔ " .
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