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Wewaii [24]
2 years ago
10

The average American man consumes 9.7 grams of sodium each day. Suppose that the sodium consumption of American men is normally

distributed with a standard deviation of 1 grams. Suppose an American man is
randomly chosen. Le X = the amount of sodium consumed. Round all numeric answers to 4 decimal places where possible.

b. Find the probability that this American man consumes between 10.5 and 11.2 grams of sodium per day? ____

c. The middle 20% of American men consume between what two weights of sodium?
Low:
High:
Mathematics
1 answer:
Phoenix [80]2 years ago
3 0

Answer:

Step-by-step explanation:

b. p(10.5<x<11.2)= p(x<11.2)-p(x<10.5)

p(x<11.2)= (11.2-9.7)/1 = 1.5 = 0.9332

p(x<10.5)= (10.5-9.7)/1= .8= 0.7881

.9332-0.7881= .1451

c. 9.7-.253, 9.7+.253

9.447, 9.953

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Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

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