Your question is: The perimeter of a rectangle is 60 meters. What are the possible lengths of any of the rectangle’s sides if its area exceeds 216 square meters?
The answer would be (12,18)
Answer:
![- \frac{2}{13}](https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B2%7D%7B13%7D%20)
Step-by-step explanation:
30w-w-6-5w-4=-5w-6-8+3w
30w-w-4=-8+3w
29w-4=-8+3w
29w-3w=-8+4
26w=-8+4
26w=-4(divide bother side by 26)
w=-2/13
If a worker makes 600,000 a month, we can multiply that to 12, since there are 12 months in a year. 600,000 x 12 = 72,000,00, so the worker makes $72,000,00 a year. :)
Answer:
![P(\bar X >70)= P(Z>\frac{70-69}{\frac{4}{\sqrt{36}}})= P(Z>1.5)](https://tex.z-dn.net/?f=%20P%28%5Cbar%20X%20%3E70%29%3D%20P%28Z%3E%5Cfrac%7B70-69%7D%7B%5Cfrac%7B4%7D%7B%5Csqrt%7B36%7D%7D%7D%29%3D%20P%28Z%3E1.5%29)
And for this case we can use the complement rule and the normal standard distribution of excel and we got:
![P(Z>1.5)=1-P(Z,1.5) = 1-0.933=0.0668](https://tex.z-dn.net/?f=%20P%28Z%3E1.5%29%3D1-P%28Z%2C1.5%29%20%3D%201-0.933%3D0.0668)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
Where
and
And we select a sample size of n =70
From the central limit theorem (n>30)we know that the distribution for the sample mean
is given by:
And we want to find this probability:
![P(\bar X >70)](https://tex.z-dn.net/?f=%20P%28%5Cbar%20X%20%3E70%29)
And we can use the z score formula given by:
![z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=%20z%3D%20%5Cfrac%7B%5Cbar%20X%20-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
And using this formula we got:
![P(\bar X >70)= P(Z>\frac{70-69}{\frac{4}{\sqrt{36}}})= P(Z>1.5)](https://tex.z-dn.net/?f=%20P%28%5Cbar%20X%20%3E70%29%3D%20P%28Z%3E%5Cfrac%7B70-69%7D%7B%5Cfrac%7B4%7D%7B%5Csqrt%7B36%7D%7D%7D%29%3D%20P%28Z%3E1.5%29)
And for this case we can use the complement rule and the normal standard distribution of excel and we got:
![P(Z>1.5)=1-P(Z,1.5) = 1-0.933=0.0668](https://tex.z-dn.net/?f=%20P%28Z%3E1.5%29%3D1-P%28Z%2C1.5%29%20%3D%201-0.933%3D0.0668)
Answer:
I not too sure if this is correct but your answer would be 737.25