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Bas_tet [7]
3 years ago
14

PLEASE HELP PLSS

Mathematics
2 answers:
Kisachek [45]3 years ago
8 0

For those who see what the person wrote above, the answer is 6.0*

bekas [8.4K]3 years ago
3 0
\cos\theta=\dfrac{\vec{u}\circ\vec{v}}{|\vec{u}|\cdot|\vec{v}|}\\\\\vec{u}= \ \textless \  2;-4 \ \textgreater \ ;\ \vec{v}= \ \textless \  3;-8 \ \textgreater \ \\\\\vec{u}\circ\vec{v}=2\cdot3+(-4)\cdot(-8)=6+32=38\\\\|\vec{u}|=\sqrt{2^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{4\cdot5}=\sqrt4\cdot\sqrt5=2\sqrt5\\\\|\vec{v}|=\sqrt{3^2+(-8)^2}=\sqrt{9+64}=\sqrt{73}\\\\|\vec{u}|\cdot|\vec{v}|=2\sqrt5\cdot\sqrt{73}=2\sqrt{365}\\\\\cos\theta=\dfrac{38}{2\sqrt{365}}\approx0.9945\to\theta\approx6^o\\\\Answer:The\ angle\ between\ \vec{u}\ and\ \vec{v}:\theta\approx6^o.
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Answer:

D. 68%

Step-by-step explanation:

The following statistics are given;

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n = 625 ; the sample size

The confidence interval for a population mean is given as;

sample mean ± z-score*\frac{s}{\sqrt{n} }

Substituting the above values we have;

47 ± z-score*\frac{5}{\sqrt{625} }

47 ± z-score*0.2

The confidence interval has been given as;

lower limit = 46.8%

upper limit = 47.2%

We can use any of these two values with the above expression to solve for the z-score. Using the lower limit we have the following equation;

47 - z-score*0.2 = 46.8

-z-score*0.2 = 46.8 - 47

-z-score*0.2 = -0.2

z-score = 1

The area of the standard normal curve between -1 and +1 will be the confidence level for the given confidence interval.

Pr( -1<Z<1 ) = 0.68 = 68%

From the Empirical rule

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4 years ago
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