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jonny [76]
3 years ago
6

Use the graph below for this question: graph of parabola going through negative 3, negative 1 and 5, negative 1. What is the ave

rage rate of change from x = −3 to x = 5?
Mathematics
1 answer:
Fofino [41]3 years ago
3 0

Here the points are(-3,-1) and (5,-1)

To find the average rate of change, we have to use the following formula .

\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Here x1 =-3, y1=-1, x2=5,y2=-1

So we get

Average rate of change=

\frac{-1+1}{5+3} = 0

So the average rate of change from x=-3 to x=5 is 0

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18 = r/3 what is r???
ohaa [14]

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3 years ago
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What will be the result of substituting 2 for x in both expressions below? One-half x + 4 x + 6 minus one-half x minus 2 Both ex
Anit [1.1K]

Answer:

Both expressions equal 5 when substituting 2 for x because both expressions are equivalent

Step-by-step explanation:

Given

\frac{1}{2}x + 4 - Expression 1

x + 6 - \frac{1}{2}x - 2 -- - Expression 2

Required

Find the result of both expressions when x = 2

<em>Expression 1</em>

\frac{1}{2}x + 4

Substitute x = 2

\frac{1}{2} * 2 + 4

1 + 4

Result = 5

<em>Expression 2</em>

x + 6 - \frac{1}{2}x - 2

Substitute x = 2

2 + 6 - \frac{1}{2} * 2 - 2

2 + 6 -1 -2

Result = 5

7 0
3 years ago
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Suppose a subdivision on the southwest side of Denver, Colorado, contains 1,500 houses. The subdivision was built in 1983. A sam
mihalych1998 [28]

Answer:

0.0268 = 2.68% probability that the sample average is greater than $229,500

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.

Single house:

The mean appraised value of a house in this subdivision for all houses is $228,000, with a standard deviation of $8,500, which means that \mu = 228, \sigma = 8.5

Sample:

Sample of 120, so, by the central limit theorem, n = 120, s = \frac{8.5}{\sqrt{120}} = 0.7759

What is the probability that the sample average is greater than $229,500?

This is 1 subtracted by the pvalue of Z when X = 229.5. So

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

By the Central Limit Theorem

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

Z = \frac{229.5 - 228}{0.7759}

Z = 1.93

Z = 1.93 has a pvalue of 0.9732

1 - 0.9732 = 0.0268

0.0268 = 2.68% probability that the sample average is greater than $229,500

7 0
3 years ago
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