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Find the value of x. A. 4 B. 6 C. 2\sqrt{3} D. 6\sqrt{3}

Bingel [31]

The answer is B hope this would help you

8 0

3 years ago

The number of people who live in a unit of area is called the population density of the area. It usually given as people "per sq

bazaltina [42]

Answer: 0.03

Step-by-step explanation:

1 block = 1/20 mile = 0.05 miles

4(0.05) = 0.20

3(0.05) = 0.15

A=l•w=0.20(0.15)=0.03

8 0

3 years ago

The amount of potato chips an 18-ounce bag contains follows a normal distribution with a mean of 18.5 ounces and a standard devi

bekas [8.4K]

Answer:

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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 question, we have that:

$\mu = 18.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02$

What is the probability that the sample mean weight of these 100 bags is less than 18.6 ounces

This is the pvalue of Z when X = 18.6. So

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

By the Central Limit Theorem

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

$Z = \frac{18.6 - 18.5}{0.02}$

$Z = 5$ has a pvalue of 1

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

7 0

3 years ago

In 1985, there were 285 cell phone subscribers in the small town of Centerville. The number of subscribers increased by 75% per

elena-s [515]

22.02............................................................................

5 0

3 years ago

A sample of 47 observations is selected from a normal population. The sample mean is 30, and the population standard deviation i

erma4kov [3.2K]

Answer:

We accept the null hypothesis and reject the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 29

Sample mean, $\bar{x}$ = 30

Sample size, n = 47

Alpha, α = 0.05

Population standard deviation, σ = 5

First, we design the null and the alternate hypothesis

$H_{0}: \mu \leq 29\\H_A: \mu > 29$

a) This is a one-tailed test because the alternate hypothesis is in greater than direction.

We use One-tailed z test to perform this hypothesis.

b) $z_{stat} > z_{critical}$ , we reject the null hypothesis and accept the alternate hypothesis and if $z_{stat} < z_{critical}$ , we accept the null hypothesis and reject the alternate hypothesis.

c) Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{30 - 29}{\frac{5}{\sqrt{47}} } = 1.37$

d) Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

Since,

$z_{stat} < z_{critical}$

We accept the null hypothesis and reject the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

e) P-value is 0.0853

On the basis of p value we again accept the null hypothesis.

5 0

4 years ago