Answer:
(0, -1)
Step-by-step explanation:
x becomes x - 6, so subtract 6 from the x-coordinate.
y becomes y - 1, so subtract 1 from the y-coordinate.
P(6, 0) -----> P'(0, -1)
Answer:
the answer is D. 5.6
Step-by-step explanation:
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You need to isolate the variable meaning .make x the only thing on one side, you do this by doing the opposite of what is being done to it. 1/2 is being added to it so subtract 1/2 from both sides of the equel sign. Next it is being multiplied by 2/3 so divide by 2/3 on both sides and you have your answer, by doing the opposite of what is being done to x you cancel everything out on that side leaving only x.
Answer:
88.88% probability that it endures for less than a year and a half
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
![\mu = 66, \sigma = 10](https://tex.z-dn.net/?f=%5Cmu%20%3D%2066%2C%20%5Csigma%20%3D%2010)
The next career begins on Monday; what is the likelihood that it endures for less than a year and a half?
One year has 52.14 weeks. So a year and a half has 1.5*52.14 = 78.21 weeks.
So this probability is the pvalue of Z when X = 78.21.
![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{78.21 - 66}{10}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B78.21%20-%2066%7D%7B10%7D)
![Z = 1.22](https://tex.z-dn.net/?f=Z%20%3D%201.22)
has a pvalue of 0.8888
88.88% probability that it endures for less than a year and a half