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vlada-n [284]
3 years ago
10

8. The endpoints of GH are G(-8, -9) and H(8,3). Find the coordinates of the midpoint M

Mathematics
2 answers:
Ad libitum [116K]3 years ago
7 0

Answer:SEE BELOW

Step-by-step explanation:

Average the X and Y coordinates (X,Y)

X = (-8 + 8)/2 = 0

Y = (-9 + 3)/2 = -3

(X,Y) = (0,-3)

dexar [7]3 years ago
7 0

Step-by-step explanation:

\underline{ \underline{ \text{Given \: points}}} :

  • G ( -8 , -9 ) & H ( 8 , 3 )

\underline{ \underline{ \text{To \: find}}} :

  • Midpoint of the given points

\underline{ \underline{ \text{Solution}}} :

Let the point G ( -8 , -9 ) be ( x₁ , y₁ ) & H ( 8 , 3 ) be ( x₂ , y₂)

\boxed{ \sf{Midpoint = (\frac{x_{1} + x_{2}}{2}} \: , \frac{y_{1} + y _{2}}{2} )}

⟶ \tt{( \frac{ - 8 + 8}{2} \:,  \frac{ - 9 + 3}{2} ) }

⟶ \tt{( \frac{0}{2}  \: , \frac{ - 6}{2}} )

⟶ \tt{(0 \:  ,- 3})

\red{ \boxed{ \boxed{ \tt{Our \: final \: answer :  \boxed{ \tt{(0 \:,  - 3)}}}}}}

Hope I helped ! ツ

Have a wonderful day / night ! ♡

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

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Answer:

3,535 cm

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Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa
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Using the z-distribution, it is found that since the <u>test statistic is greater than the critical value</u>, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is <u>tested if the mean length of jail time is still of 2.5 years</u>, that is:

H_0: \mu = 2.5

At the alternative hypothesis, it is <u>tested if it has increased</u>, that is:

H_1: \mu > 2.5

We have the <u>standard deviation for the population</u>, thus, the z-distribution is used. The test statistic is given by:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • \sigma is the standard deviation of the sample.
  • n is the sample size.

For this problem, the values of the <u>parameters</u> are: \overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26

Hence, the value of the <u>test statistic</u> is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}

z = 1.7

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Since the <u>test statistic is greater than the critical value</u>, it can be concluded that the mean length of jail time has increased.

A similar problem is given at brainly.com/question/24166849

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An arithmetic progression is simply a progression with a common difference among consecutive terms.

  • <em>The sum of multiplies of 6 between 8 and 70 is 390</em>
  • <em>The sum of multiplies of 5 between 12 and 92 is 840</em>
  • <em>The sum of multiplies of 3 between 1 and 50 is 408</em>
  • <em>The sum of multiplies of 11 between 10 and 122 is 726</em>
  • <em>The sum of multiplies of 9 between 25 and 100 is 567</em>
  • <em>The sum of the first 20 terms is 630</em>
  • <em>The sum of the first 15 terms is 480</em>
  • <em>The sum of the first 32 terms is 3136</em>
  • <em>The sum of the first 27 terms is -486</em>
  • <em>The sum of the first 51 terms is 2193</em>

<em />

<u>(a) Sum of multiples of 6, between 8 and 70</u>

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

\mathbf{a = 12}

\mathbf{n = 10}

\mathbf{d = 6}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}

\mathbf{S_{10} = 390}

<u>(b) Multiples of 5 between 12 and 92</u>

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

\mathbf{a = 15}

\mathbf{n = 16}

\mathbf{d = 5}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}

\mathbf{S_{16} = 840}

<u>(c) Multiples of 3 between 1 and 50</u>

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

\mathbf{a = 3}

\mathbf{n = 16}

\mathbf{d = 3}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}

\mathbf{S_{16} = 408}

<u>(d) Multiples of 11 between 10 and 122</u>

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

\mathbf{a = 11}

\mathbf{n = 11}

\mathbf{d = 11}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}

\mathbf{S_{11} = 726}

<u />

<u>(e) Multiples of 9 between 25 and 100</u>

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

\mathbf{a = 27}

\mathbf{n = 9}

\mathbf{d = 9}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}

\mathbf{S_{9} = 567}

<u>(f) Sum of first 20 terms</u>

The given parameters are:

\mathbf{a = 3}

\mathbf{d = 3}

\mathbf{n = 20}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}

\mathbf{S_{20} = 630}

<u>(f) Sum of first 15 terms</u>

The given parameters are:

\mathbf{a = 4}

\mathbf{d = 4}

\mathbf{n = 15}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}

\mathbf{S_{15} = 480}

<u>(g) Sum of first 32 terms</u>

The given parameters are:

\mathbf{a = 5}

\mathbf{d = 6}

\mathbf{n = 32}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}

\mathbf{S_{32} = 3136}

<u>(g) Sum of first 27 terms</u>

The given parameters are:

\mathbf{a = 8}

\mathbf{d = -2}

\mathbf{n = 27}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}

\mathbf{S_{27} = -486}

<u>(h) Sum of first 51 terms</u>

The given parameters are:

\mathbf{a = -7}

\mathbf{d = 2}

\mathbf{n = 51}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}

\mathbf{S_{51} = 2193}

Read more about arithmetic progressions at:

brainly.com/question/13989292

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2 years ago
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