$\frac{\sin B}{b}=\frac{\sin A}{a} \\ \\ \sin B=\frac{b \sin A}{a} \\ \\ \sin B=\frac{13 \sin 29^{\circ}}{7.2} \\ \\ \boxed{m\angle B =\arcsin \left(\frac{65\sin 29^{\circ}{36} \right), 180^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$

Considering the diagram. $\angle B$ is shown to be acute, so $\boxed{m\angle B=\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)}$ .

Angles in a triangle add to 180°, so $m\angle C= 151^{\circ}-\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right)$ .

Using the law of cosines,

$AB=\sqrt{13^2 + 7.2^2-2(13)(7.2)\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)} \\ \\ \boxed{AB=\sqrt{220.84-187.2\cos \left(\arcsin \left(\frac{65\sin 29^{\circ}}{36} \right) \right)}}$

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1 year ago

A normal population has a mean of 19 and a standard deviation of 5.

dangina [55]

Answer:

a) $Z = 1.2$

b) 38.49% of the population is between 19 and 25.

c) 34.46% of the population is less than 17.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. If we need to find the probability that the measure is larger than X, it is 1 subtracted by this pvalue.

For this problem, we have that

A normal population has a mean of 19 and a standard deviation of 5, so $\mu = 19, \sigma = 5$ .