Factor 49a– 14a +1.<br> 49a

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ... No; The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: No; 9.

8 0

3 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0

3 years ago

Graph the first six terms of a sequence where a1=5 and r=1.25

emmainna [20.7K]

The sequence above is geometric progression.
The nth term of such sequence is given by;

Tn = ar∧(n-1),
Where a⇒first term and
r⇒common ratio

So, 1st term = 5×1.25∧(1-1) = 5×1.25∧0 =5
2nd term = 5×1.25∧(2-1) = 5×1.25 = 6.25
3rd term = 5×1.25∧(3-1) = 5×1.25² = 7.8125
4th term = 5×1.25∧(4-1) =5×1.25³ = 9.765625
5th term = 5×1.25∧(5-1) = 5×1.25∧4 = 12.20703125
6th term = 5×1.25∧(6-1) = 5×1.25∧5 = 15.25878909

7 0

3 years ago

Factor 49a– 14a +1. 49a– 14a + 1 =