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3 years ago

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

Mathematics

2 answers:

igomit [66]3 years ago

8 0

Answer:

x = 88

A = 94°

B = 102°

C = 86°

D = 78°

Step-by-step explanation:

We find the angle measurements by finding x first.

what we first do in this equation x+6+x-2=180 is combine like terms

this would equal

2x+4=180

we want x by itself so we subtract 4 from both sides

this would equal

2x=176

x is still not by itself so now we divide 2 by both sides

this would equal

x = 88

Now we just plug in our answers in to the angles meaning

we replace everything that is x in the quadrilateral with 88

Now that we found all the sides with equations there is still one angle that is undefined

We figure this out by subtracting all the angles with 360.

this is because all the angles degrees added in a quadrilateral equals 360. so 360-258=102

so the angle measurement of angle B is 102

kherson [118]3 years ago

4 0

Answer:

x=88

Explanation:

x+6+x-2=180

2x+4=180

2x=180-4

2x=176

*divivd both sides*

x=88

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At one gym, there is a $12 start-up fee, and after that each month at the gym costs $20. At another gym, it costs $4 to startup,

mart [117]

Answer:

In 4 months the cost of both gyms will be the same.

Step-by-step explanation:

At first we need to model the function to calculate the cost of the 2 gyms.

Slope-intercept equation of linear function

$f(x)=mx+b$

where

$m\rightarrow$ slope of line

$b\rightarrow$ y-intercept

Let linear function to calculate total cost of gym be:

$c(n)=mn+b$

where

$c\rightarrow$ total cost of gym

$m\rightarrow$ cost per month (slope)

$n\rightarrow$ number of months

$b\rightarrow$ start-up fee (y-intercept)

For Gym 1

$m=20$ , $b=12$

$c(n)=20n+12$

For Gym 2

$m=22$ , $b=4$

$c(n)=22n+4$

In order to find the number of months the cost of both gyms will be the same, we need to equate both functions and solve for number of months $(n)$

$20n+12=22n+4$

$\\\textrm{Subtracting 22n from both sides}\\20n-22n+12=22n-22n+4\\-2n+12=4\\\textrm{Subtracting 12 from both sides}\\-2n+12-12=4-12\\-2n=-8\\\textrm{Dividing both sides by -2}\\\frac{-2}{-2}n=\frac{-8}{-2}\\n=4\textrm{ months}$

So,

In 4 months the cost of both gyms will be the same.

5 0

4 years ago

Please help mehhh Thank you! :)

vivado [14]

Parentheses first -4+-8= -12(1)= -12
= -12

4 0

4 years ago

(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot

VladimirAG [237]

Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

We could model this situation with a binomial distribution

$P(6;k)=\binom{6}{k}p^kq^{6-k}$

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

$\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}$

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

$P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}$

$P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281$

$P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484$

$P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438$

$P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109$

$P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781$

and adding up these values we have that the probability that Y is less than 6 is

$\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%$

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.

In a binomial distribution, the standard deviation is

$s = \sqrt{npq}$

In this case

$\sqrt{200(0.85)(0.15)}=5.05$

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

$np\geq 5 \;and\;nq \geq 5$