How to solve 3/4 = 8/x-1

if α and β are the zeroes of the polynomial 3x^2 - 8x − 3 then find the value of i) α^2 + β^2 ii) α^-1 + β^-1

BlackZzzverrR [31]

Step-by-step explanation:

the answer is in the above image

5 0

3 years ago

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4. Suppose 64 golfers play

Vilka [71]

Answer:

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

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:

$\mu = 75, \sigma = 4, n = 64, s = \frac{4}{\sqrt{64}} = 0.5$

Find the probability that the average score of the 64 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 64.

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

By the Central Limit Theorem

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

$Z = \frac{76 - 75}{0.5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

6 0

4 years ago

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour

Lesechka [4]

Answer:

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=54$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=21$ represent the sample standard deviation

n=16 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$