Answer:
My answers were (1,1) for both lmk if it works ;)
Step-by-step explanation:
Answer:
Step-by-step explanation:
![(\sqrt[4]{9})^{\frac{1}{2}x}=(9^{\frac{1}{4}})^{\frac{1}{2}x}\\\\=9^{\frac{1}{4}*\frac{1}{2}x}\\\\=9^{\frac{1}{8}x}](https://tex.z-dn.net/?f=%28%5Csqrt%5B4%5D%7B9%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D%3D%289%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D%5C%5C%5C%5C%3D9%5E%7B%5Cfrac%7B1%7D%7B4%7D%2A%5Cfrac%7B1%7D%7B2%7Dx%7D%5C%5C%5C%5C%3D9%5E%7B%5Cfrac%7B1%7D%7B8%7Dx%7D)
Answer:
Y - intercept is 1 , slope is 3
Answer:

Step-by-step explanation:
The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.
To find the
percentile for the television weights, use the formula:
, where
is the average of the set,
is some constant relevant to the percentile you're finding, and
is one standard deviation.
As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute
,
, and
:

Therefore, the 90th percentile weight is 5.1282 pounds.
Repeat the process for calculating the 10th percentile weight:

The difference between these two weights is
.