320/240 = 1.33
1.33 = 133% (move the decimal point over two places to the right)
133% is your answer
hope this helps
Answer: 0.0386
Step-by-step explanation:
Given: The population of 400 tall women has a mean height
of 179.832 cm and a standard deviation
of 12.192 cm.
Let X be a random variable that represents the height of woman.
Sample size : n= 50
The probability that the mean for this sample group is above 182.88 will be :
![P(\overline{X}>182.88)\\\\=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{182.88-179.832}{\dfrac{12.192}{\sqrt{50}}})\\\\ =P(Z>1.7678)\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Coverline%7BX%7D%3E182.88%29%5C%5C%5C%5C%3DP%28%5Cdfrac%7B%5Coverline%7BX%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%3E%5Cdfrac%7B182.88-179.832%7D%7B%5Cdfrac%7B12.192%7D%7B%5Csqrt%7B50%7D%7D%7D%29%5C%5C%5C%5C%20%3DP%28Z%3E1.7678%29%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7B%5Coverline%7BX%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C1.7678%29%5C%5C%5C%5C%3D1-0.9614%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20p-value%20table%7D%5D%5C%5C%5C%5C%3D%200.0386)
Hence, Required probability = 0.0386
Answer:
m<4= 60
m<1=60
m<3=30
Step-by-step explanation:
I'd like to explain but I would be very difficult to do with out being in person
Answer:
27
Step-by-step explanation:
Use synthetic division:
8 | 1 -5 3
__8_24
1 3 | 27
Therefore, (x²-5x+3)/(x-8) = x+3 R27