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3 years ago

Quadrilateral ABCD is inscribed in this circle.

Mathematics

2 answers:

Alecsey [184]3 years ago

8 0

The sum of all interior angles in a polygon is

180(n-2), where n = sides, well, this is a QUADrilateral, so it has 4 sides, so the total is 360°.

now, let's find what angle C is first,

$\bf \stackrel{A}{(2x-40)}+\stackrel{B}{(116)}+C+\stackrel{D}{(x)}=360\implies C+3x+76=360
\\\\\\
C+3x=284\implies C=284-3x$

now, recall the "inscribed quadrilateral conjecture", where opposite angles are "supplementary angles", thus

$\bf \stackrel{\measuredangle A}{(2x-40)}+\stackrel{\measuredangle C}{(284-3x)}=180\implies -x+244=180
\\\\\\
64=x\\\\
-------------------------------\\\\
\measuredangle A=2(64)-40$

victus00 [196]3 years ago

6 0

The answer should be 88

Hope this helps and have good day

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95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

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We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of residents 65-years or older who had obtained a flu shot = $\frac{28}{111}$ = 0.25

n = sample of residents 65-years or older = 111

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Here for constructing 95% confidence interval we have used One-sample z test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ , $0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ ]