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statuscvo [17]
3 years ago
9

Find m∠1 if m∠2=73°, m∠3=107°, m∠4=92°. Justify your response!

Mathematics
1 answer:
a_sh-v [17]3 years ago
7 0

Answer:

m∠1 = 92°

Step-by-step explanation:

∠2 corresponds to the supplement of ∠3 if (and only if) lines a and b are parallel. We find that

m∠2 + m∠3 = 73° +107° = 180°

so, the angles are supplementary and lines a and b are parallel.

Angles 4 and 1 are corresponding angles where the line d crosses the parallel lines a and b, so are congruent.

m∠1 = m∠4 = 92°

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What is the domain of the function f(x)= e^x/e^x+c if c is a constant greater than 0
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La suma de tres números enteros consecutivos es 156 halla los números
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(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
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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:

a)

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.

So

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

b)  

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\%

c)

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

in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30

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