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Nataly_w [17]
3 years ago
11

We say that an integer a is a type 0 integer if there exists an integer n such that a = 3n. An integer a is a type 1 integer if

there exists an integer n such that a = 3n + 1. An integer a is a type 2 integer if there exists an integer n such that a = 3n + 2. Prove that if a is a type 1 integer, then a 2 is a type 1 integer\
Mathematics
2 answers:
Delicious77 [7]3 years ago
5 0

Answer:

<em>Proof below</em>

Step-by-step explanation:

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove a^2 is a type 1 integer

Expanding

a^2=(3n+1)^2=9n^2+6n+1

If a^2 is a type 1 integer, then we should be able to find an integer m such as

a^2=3m+1

Equating

a^2=3m+1=9n^2+6n+1

solving for m

m=3n^2+2n

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

Gnoma [55]3 years ago
5 0

hope this helps :)

have an amazing day :)

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove  is a type 1 integer

Expanding

If  is a type 1 integer, then we should be able to find an integer m such as

Equating

solving for m

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

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Scores on a college entrance exam are normally distributed with a mean of 550 and a standard deviation of 100. Find the value th
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Answer:

The value that represents the 90th percentile of scores is 678.

Step-by-step explanation:

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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:

\mu = 550, \sigma = 100

Find the value that represents the 90th percentile of scores.

This is the value of X when Z has a pvalue of 0.9. So X when Z = 1.28.

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1.28 = \frac{X - 550}{100}

X - 550 = 100*1.28

X = 678

The value that represents the 90th percentile of scores is 678.

4 0
2 years ago
Solve the equation <img src="https://tex.z-dn.net/?f=x%5E2%2B25%3D0" id="TexFormula1" title="x^2+25=0" alt="x^2+25=0" align="abs
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Step-by-step explanation:

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Rewriting

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Writing as the difference of squares

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