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2 years ago

7

If a jar contains 3 blueberry, 4 ginger, 1 green, and 6 regular tea bags, what is the probability that a regular tea bag will be

pulled from the jar? Answers must be in simplest form.

1 answer:

Nezavi [6.7K]2 years ago

3 0

I love blueberries so 10+y is not right answer

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3 years ago

Factor the polynomial. 6g8 – 3g4 + 9g2 urgentt

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3 years ago

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2 years ago

Given f(x) = 17-xwhat is the average rate of change in f(x) over the interval [1, 5]?

stira [4]

<h2>Answer:</h2>

The average rate of change in f(x) over the interval [1, 5] is -1

<h2>Step-by-step explanation:</h2>

Hi! Let me help you to understand this problem. Here we have the following function:

$f(x) = 17-x$

We need to compute the Average Rate of Change (ARC) in $f(x)$ over the interval $[1, 5]$ . So <em>what is the average rate of change of a function? </em>In general, for a nonlinear graph whose slope changes at each point, the average rate of change between any two points $(x_{1},f(x_{1}) \ and \ (x_{2},f(x_{2})$ is defined as the slope of that line through that two points. Here we have a linear function, so the average rate of change will be the slope of the line:

So:

$ARC=m=-1$

This can also be calculated as:

$ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} \\ \\ ARC=\frac{17-5-(17-1)}{5-1} \\ \\ ARC=-1$

4 0

3 years ago

A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Qua

amid [387]

Answer:

There is a 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be 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.

In this problem, we have that:

The population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch. This means that $\mu = 120, \sigma = 0.05$ .

Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. This means that $n = 100, s = \frac{\sigma}{\sqrt{n}} = \frac{0.05}{\sqrt{100}} = 0.005$ .

The probability that Claude's sample has a mean between 119.985 and 120.0125 inches is

We are working with the sample mean, so we use the standard deviation of the sample, that is, $s$ instead of $\sigma$ in the z score formula.

This probability is the pvalue of Z when $X = 120.0125$ subtracted by the pvalue of Z when $X = 119.985$ .

X = 120.0125

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

$Z = \frac{120.0125 - 120}{0.005}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938.

X = 119.985

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

$Z = \frac{119.985 - 120}{0.005}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0014.

So there is a 0.9938 - 0.0014 = 0.9924 = 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

7 0

3 years ago