What's the unit rate for $44 and 4 days

Mathematics

2 answers:

djverab [1.8K]4 years ago

6 0

I think that would be $11 a day because $44 divided by 4 days is $11 a day.

iren2701 [21]4 years ago

3 0

11
i think because to find a unit rate you divide the two numbers

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Problem 2. A random variable has the following density function:

kolbaska11 [484]

If <em>f(x)</em> is a density function, then its integral over the entire real line should be 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_0^2(1-0.5x)\,\mathrm dx=(x-0.25x^2)\bigg|_0^2=2-1=1$

so yes, <em>f(x)</em> is indeed a density function.

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3 years ago

A population of plastic chairs in a factory has a weight's mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample

Firlakuza [10]

Answer:

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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:

$\mu = 1.5, \sigma = 0.1, n = 100, s = \frac{0.1}{\sqrt{100}} = 0.01$

What is the probability that the sample mean will be within +0.02 of the population mean?

Sample mean between 1.5 - 0.02 = 1.48 kg and 1.5 + 0.02 = 1.52 kg, which is the pvalue of Z when X = 1.52 subtracted by the pvalue of Z when X = 1.48. So

X = 1.52

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

By the Central Limit Theorem

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