Answer: B
Step-by-step explanation:
The 15 is on the y-axis and it’s positive so it goes up
Answer:
yes
Step-by-step explanation:
The probability of a normally distributed data with mean, μ and standard deviation, σ, exceeds a value, a, is given by
![P(x \ \textgreater \ a) = 1 - P(x \leq a) = 1 - P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)](https://tex.z-dn.net/?f=P%28x%20%5C%20%5Ctextgreater%20%5C%20%20a%29%20%3D%201%20-%20P%28x%20%5Cleq%20a%29%20%3D%201%20-%20P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%20%5Cfrac%7Ba-%5Cmu%7D%7B%5Csigma%7D%20%5Cright%29)
<span>Given that the
average yearly snowfall in chillyville is normally distributed with a
mean of 55 inches. If the snowfall in chillyville exceeds 60 inches in
15% of the years, thus we have:
![1 - P\left(z\ \textless \ \frac{60-55}{\sigma} \right)=0.15 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{5}{\sigma} \right)=1-0.15=0.85 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{5}{\sigma} \right)=P(z\ \textless \ 1.036) \\ \\ \Rightarrow \frac{5}{\sigma}=1.036 \\ \\ \Rightarrow\sigma= \frac{5}{1.036} =4.83](https://tex.z-dn.net/?f=1%20-%20P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7B60-55%7D%7B%5Csigma%7D%20%5Cright%29%3D0.15%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7B5%7D%7B%5Csigma%7D%20%5Cright%29%3D1-0.15%3D0.85%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7B5%7D%7B%5Csigma%7D%20%5Cright%29%3DP%28z%5C%20%5Ctextless%20%5C%201.036%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cfrac%7B5%7D%7B%5Csigma%7D%3D1.036%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Csigma%3D%20%5Cfrac%7B5%7D%7B1.036%7D%20%3D4.83)
Therefore, the standard deviation is approximately 5.
</span>
3x-14=4x-9
-4x -4x
-x-14=-9
-14 -14
-x=-15
-x/-1=-15/-1
x=15
Sorry if this is a bit confusing, but I hope I helped