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vovikov84 [41]
2 years ago
7

Teresa is factoring this polynomial by grouping. Which common factors should be used in the next step of factoring? 10x3 3x2−20x

−6 (10x3 3x2) (−20x−6) x2 and −2x 2x2 and −2x x2 and −2 2x2 and −2.
Mathematics
2 answers:
Oksi-84 [34.3K]2 years ago
8 0

Answer:

−2=−2x^3

Step-by-step explanation:

10x3= 3x2−20x−6=(10x3⋅(3x2))(−20x−6)=−600x6−180x5x2=−600x6−180x5−2x⋅(2x2)=−4x3−2x⋅x2=−2x^3

−2=−2x3

2x^2=−2x^3

−2=−2x^3

Elanso [62]2 years ago
4 0

Answer:

> The correct choice is C) x2 and −2

Step-by-step explanation:

> This is the only answer choice that makes sense in this equation.

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Read 2 more answers
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

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Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

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3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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