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Aleks [24]
3 years ago
10

Solve the following system of equations using any method you want.

Mathematics
2 answers:
aniked [119]3 years ago
8 0

Answer:

y = - 2, x = 2

Step-by-step explanation:

Remove x and solve for y

2x - 2x - 5y - y = 12

-6y = 12

y = - 2

Substitute y back into the equation

-2 + 2x = 2

2x = 4

x = 2

Daniel [21]3 years ago
3 0
2x+y=2
2x-5y=14
Subtract the two equation
6y= -12
Y= - 2
Put y= -2 in any equation
2x-2=2
2x=4
X=2
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10................/......
jolli1 [7]

Answer:

<h2><u>-1 + √3 or -(1 - 2√3)</u></h2>

Step-by-step explanation:

(1 + √3) (2 - √3) = 2 - √3 + 2√3 - 3 = 2 - 3 - √3 + 2√3 = <u>-1 + √3 or -(1 - 2√3)</u>

6 0
3 years ago
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3y^4/3y^2-6=10 please help I will.mark it as the brainliest answer!​
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Y=4 or -4
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3 years ago
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Chris made his aunt a garden. The width is 7 2/5 ft and the length is 5 1/2 ft. What is the area of the garden?
pishuonlain [190]

Answer:

Hello! if Chris garden is a square shape then the answer is 40 7/10 in fraction form or 40.7 in decimal form

6 0
2 years ago
The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean
Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51%  probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

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 problem, we have that:

\mu = 591, \sigma = 42

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 591}{42}

Z = 0.21

Z = 0.21 has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have n = 50, s = \frac{42}{\sqrt{50}} = 5.94

This is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 591}{5.94}

Z = 1.515

Z = 1.515 has a pvalue of 0.9351

93.51%  probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have n = 50, s = \frac{42}{\sqrt{100}} = 4.2

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

Z = \frac{600 - 591}{4.2}

Z = 2.14

Z = 2.14 has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

8 0
3 years ago
Solve for a a=.5×12ft (30ft+42ft)
Anastasy [175]
.5x12 (30+42)
    v          v
    6    *    72= 432 ft
7 0
3 years ago
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