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rusak2 [61]
3 years ago
9

Can you guys graph a system of equations and make sure it has two parabolas with the same vertex and a line that crosses the sam

e vertex of the 2 parabolas. Thanks will mark brainliest

Mathematics
2 answers:
Andrews [41]3 years ago
6 0

Answer:

Sketch the following three graphs:

1) y = (x - 5)²

2) y = -(x - 5)²

3) y = x - 5

First two are parabolas with vertex at (5,0)

Third one is a line which intersects both the parabolas as (5,0)

Andrej [43]3 years ago
5 0

Step-by-step explanation:

<u>Step 1:  Make a system of equations</u>

The easiest place to place the parabolas are at the origin or (0, 0).  So, the parabolas are going to have an equation similar to y = x^2.  The line that crosses the same vertex of the 2 parabolas is going to be y = 0 because both of the parabolas vertex is going to be at (0, 0).

y = 0 is the line

y = x^2 is one of the parabolas

y = 2x^2 is another one of the parabolas

Look at the graph below

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Find the Unit Rate: If 10 30.5 oz containers cost $70 how much does one container cost?​
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(N^3 - N^4) - (3n^3- 7n^4)
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If you would like to solve (n^3 - n^4) - (3n^3 - 7n^4), you can do this using the following steps:

(n^3 - n^4) - (3n^3 - 7n^4) = <span>n^3 - n^4 - 3n^3 + 7n^4 = n^3 - 3n^3 - n^4 + 7n^4 = -2n^3 + 6n^4
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Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

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2 years ago
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