2. Quadrilateral ABCD is inscribed in a circle. Find the measure of each of the angles of the quadrilateral. Show your work

Mathematics

1 answer:

Nostrana [21]3 years ago

7 0

On this picture is shown a quadrilateral inscribed in a circle and by the Inscribed Quadrilateral Theorem the angles on the opposite vertices are supplementary, or in other words are equals to 180 degrees.

On this exercise it is asked to find the measure of angle B, First of all, you need to find the value of x. To so you have to select two opposite angles on this case angles A and C.

m<A+m<C=180 Substitute the given values for angles A and C
x+2+x-2=180 Combine like terms
2x=180 Divide by 2 in both sides to isolate x
x=90

Now, that the value of x is known you can substitute it in the expression representing angle D, and then subtract that number from 180 to find the measure of angle B.

m<D=x-10 Substitute the value of x
m<D=90-10 Combine like terms
m<D=80

m<B=180-m<D Substitute the value of angle D
m<B=180-80 Combine like terms
m<B=100

The measure of angle B is 100 degrees, and the value of x is 90.

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The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ 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 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95