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koban [17]
3 years ago
9

Given points A (-3,-4) and B (2,0), points P (-1,-12/5) partitions line AB in the ratio

Mathematics
1 answer:
kvv77 [185]3 years ago
7 0
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
&A&(~ -3 &,& -4~) 
&P&(~ -1 &,& -\frac{12}{5}~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AP=\sqrt{[-1-(-3)]^2+[-\frac{12}{5}-(-4)]^2}
\\\\\\
AP=\sqrt{(-1+3)^2+(-\frac{12}{5}+4)^2}\implies AP=\sqrt{2^2+\left( \frac{8}{5} \right)^2}
\\\\\\
AP=\sqrt{4+\frac{64}{25}}\implies AP=\sqrt{\cfrac{164}{25}}\implies AP=\cfrac{\sqrt{164}}{\sqrt{25}}
\\\\\\
\boxed{AP=\cfrac{2\sqrt{41}}{5}}



\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
&P&(~ -1 &,& -\frac{12}{5}~) 
&B&(~ 2 &,& 0~)
\end{array}
\\\\\\
PB=\sqrt{[2-(-1)]^2+[0-\left(-\frac{12}{5}  \right)]^2}
\\\\\\
PB=\sqrt{(2+1)^2+\left( \frac{12}{5} \right)^2}\implies PB=\sqrt{9+\frac{144}{25}}\implies PB=\sqrt{\cfrac{369}{25}}
\\\\\\
PB=\cfrac{\sqrt{369}}{\sqrt{25}}\implies \boxed{PB=\cfrac{3\sqrt{41}}{5}}

so, let's check the ratio of AP:PB

\bf AP:PB\implies \cfrac{AP}{PB}\implies \cfrac{\frac{2\sqrt{41}}{5}}{\frac{3\sqrt{41}}{5}}\implies \cfrac{2\underline{\sqrt{41}}}{\underline{5}}\cdot \cfrac{\underline{5}}{3\underline{\sqrt{41}}}\implies \cfrac{2}{3}
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GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

5 0
3 years ago
Please help help help help please help will give brainliest if correct​
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Answer:

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Step-by-step explanation:

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6 divided by 3 = 2

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So 1/3 is the smaller fraction, 7/12 is in the middle and 2/3 is the bigger fraction.
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