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3 years ago

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Classify the sequence {an) = {10,4,8/5,16/25,...} as arithmetic, geometric, or neither. If there is not enough information to cl

assify the sequence, chose not enough information.
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-
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A. arithmetic
B. geometric
C. neither
D. not enough information

1 answer:

Vanyuwa [196]3 years ago

4 0

The answer is arithmetic

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Which is an equivalent expression for 4(x + 7) – 9

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Answer:

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Step-by-step explanation:

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Select 3 expressions that have a sum or difference of 3 /4 .

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Answer:

The Three expressions that have a sum or difference of 3 /4 .

B . 11/12 − 1/6

C. 3/5 + 3/20

E . 2/3 + 1/12

Step-by-step explanation:

The Three expressions that have a sum or difference of 3 /4 .

are

B . 11/12 − 1/6

$\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{11}{12}-\dfrac{1\times 2}{6\times 2}\\\\\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}$

Therefore,

$\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{3}{4}$

C. 3/5 + 3/20

$\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{3\times 4}{5\times 4}+\dfrac{3}{20}\\\\\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{12}{20}+\dfrac{3}{20}=\dfrac{15}{20}=\dfrac{3}{4}$

Therefore,

$\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{3}{4}$

E . 2/3 + 1/12

$\dfrac{2}{3}+\dfrac{1}{12}=\dfrac{2\times 4}{3\times 4}+\dfrac{1}{12}\\\\\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{8+1}{12}=\dfrac{9}{12}=\dfrac{3}{4}$

Therefore,

$\dfrac{2}{3}+\dfrac{1}{12}=\dfrac{3}{4}$

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3 years ago

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3 years ago

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that

kompoz [17]

Answer:

a)

<em>The probability that a single score drawn at random will be greater than 110 </em>

<em>P( X > 110) = 0.1587</em>

<em>b) </em>

<em>The probability that a sample of 25 scores will have a mean greater than 105</em>

<em> P( x> 105) = 0.0062</em>

<em>c) </em>

<em>The probability that a sample of 64 scores will have a mean greater than 105</em>

<em> P( x⁻> 105) = 0.002</em>

<em></em>

<em>d) </em>

<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>

<em> P( 95 ≤ X≤ 105) = 0.9544</em>

<em></em>

Step-by-step explanation:

<u><em>a)</em></u>

Given mean of the Normal distribution 'μ' = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

<em>The probability that a single score drawn at random will be greater than 110</em>

<em>P( X > 110) = P( Z >1)</em>

= 1 - P( Z < 1)

= 1 - ( 0.5 +A(1))

= 0.5 - A(1)

= 0.5 -0.3413

= 0.1587

b)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

<em>The probability that a single score drawn at random will be greater than 110</em>

<em> P( x> 105) = P( z > 2.5)</em>

<em> = 1 - P( Z< 2.5)</em>

<em> = 1 - ( 0.5 + A( 2.5))</em>

<em> = 0.5 - A ( 2.5)</em>

<em> = 0.5 - 0.4938</em>

<em> = 0.0062</em>

<em>The probability that a single score drawn at random will be greater than 105</em>

<em> P( x> 105) = 0.0062</em>

<em>c) </em>

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

<em>The probability that a single score drawn at random will have a mean greater than 105</em>

<em> P( x> 105) = P( z > 4)</em>

<em> = 1 - P( Z< 4)</em>

<em> = 1 - ( 0.5 + A( 4))</em>

<em> = 0.5 - A ( 4)</em>

<em> = 0.5 - 0.498</em>

<em> = 0.002</em>

<em> The probability that a sample of 64 scores will have a mean greater than 105</em>

<em> P( x⁻> 105) = 0.002</em>

<em>d) </em>

<em>Let x₁ = 95</em>

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

<em>Let x₂ = 105</em>

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

= P(z≤2) - P(z≤-2)

= 0.5 + A( 2) - ( 0.5 - A( -2))

= A( 2) + A(-2) (∵A(-2) =A(2)

= A( 2) + A(2)

= 2 × A(2)

= 2×0.4772

= 0.9544

<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>

<em> P( 95 ≤ X≤ 105) = 0.9544</em>

<em> </em>

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3 years ago