Volume of a shape

Mujer culona

KIM [24]

Mujer culona mujer culona mujer culona

6 0

3 years ago

For questions 8 – 10, answer the questions about tangent-tangent angles.

olchik [2.2K]

Answer:

m(arc ZWY) = 305°

Step-by-step explanation:

8). Formula for the angle formed outside the circle by the intersection of two tangents or two secants is,

Angle formed by two tangents = $\frac{1}{2}(\text{Difference of intercepted arcs})$

= $\frac{1}{2}(220-140)$

= $\frac{1}{2}(80)$

= 40°

9). Following the same rule as above,

Angle formed between two tangents = $\frac{1}{2}(\text{Difference of intercepted arcs})$

125 = $\frac{1}{2}[m(\text{major arc})-m(\text{minor arc})]$

250 = $[m(\text{arc ZWY})-m(\text{arc ZY})]$

250 = m(arc ZWY) - 55

m(arc ZWY) = 305°

Therefore, measure of arc ZWY = 305° will be the answer.

10). m(arc BAC) = $\frac{1}{2}([m(\text{arc BDC})-m(\text{arc BC})])$

= $\frac{1}{2}(254-106)$

= $\frac{148}{2}$

= 74°

4 0

2 years ago

Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr

bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .