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kati45 [8]
4 years ago
15

What is the answer for 6r^2-10r-24

Mathematics
1 answer:
FrozenT [24]4 years ago
8 0
6r^2-10r-24 =0\\ \\a=6, \ b= -10, \ c= -24 \\ \\ \Delta =b^2-4ac =  (-10)^2 -4\cdot6\cdot  (-24) =  100+576 = 676 \\ \\ r_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{10-\sqrt{676}}{2\cdot 6 }=\frac{ 10-26}{12}= \frac{-16}{12}=- \frac{4}{3} \\ \\r_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{10+\sqrt{676}}{2\cdot 6 }=\frac{ 10+26}{12}= \frac{36}{12}=3\\ \\   6r^2-10r-24  = 6(r+\frac{4}{3})(r-3)=(6r+8)(r-3)\\ \\ \\ Answer : \ 6r^2-10r-24 =(6r-8)(r-3)
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What proportion of US women have a height greater than 69.5 inches?
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Using the Normal distribution, it is found that 0.0359 = 3.59% of US women have a height greater than 69.5 inches.

In a <em>normal distribution</em> with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It measures how many standard deviations the measure is from the mean.  
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US women’s heights are normally distributed with mean 65 inches and standard deviation 2.5  inches, hence \mu = 65, \sigma = 2.5.

The proportion of US women that have a height greater than 69.5 inches is <u>1 subtracted by the p-value of Z when X = 69.5</u>, hence:

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

Z = \frac{69.5 - 65}{2.5}

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

1 - 0.9641 = 0.0359

0.0359 = 3.59% of US women have a height greater than 69.5 inches.

You can learn more about the Normal distribution at brainly.com/question/24663213

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3 years ago
Identify the range of the functions shown in the graph
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List of numbers Irrational and suspected irrational numbers
γ ζ(3) √2 √3 √5 φ ρ δS e π δ
Binary 10.0011110001101110…
Decimal 2.23606797749978969…
Hexadecimal 2.3C6EF372FE94F82C…
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2
+
1
4
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4
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⋱
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5
.
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1
/
10,000
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