Also this one! :( pls help again! thanks.

A triangle has 180 degrees as a whole. An obtuse angle is an angle that has more than 90 degrees while a right right angle has 90 degrees. So, if were to be added, it will give a result of more than 180 degrees. So, the only angle can be added to a triangle that has an obtuse angle is an acute angle, since an acute is an angle that has less than 90 degrees.

4 0

3 years ago

Suppose a batch of steel rods produced at a steel plant have a mean length of 170 millimeters, and a standard deviation of 10 mi

iragen [17]

Answer:

77.38% probability that the mean length of the sample rods would differ from the population mean by less than 0.7 millimeters

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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:

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

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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 170, \sigma = 10, n = 299, s = \frac{10}{\sqrt{299}} = 0.5783$

If 299 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by less than 0.7 millimeters

This is the pvalue of Z when X = 170 + 0.7 = 170.7 subtracted by the pvalue of Z when X = 170 - 0.7 = 169.3. So

X = 170.7

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

By the Central Limit Theorem

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