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3 years ago

Solve the system of equations x+y=-8 and -9x-6y=60 by combining the equations

Mathematics

1 answer:

TiliK225 [7]3 years ago

8 0

Answer:

x = - 4 y = - 4

Step-by-step explanation:

x+y= - 8

-9x-6y=60

First, solve for x in the first equation:

x+y = - 8 Subtract y from both sides

x + y - y = -8 - y y cancels on the left

x = - 8 - y

Now plug in what you found for x into the 2nd equation and solve for y.

- 9x - 6y = 60

-9(- 8 - y) - 6y = 60 Multiply out

72 + 9y - 6y = 60

72 + 3y = 60 Subtract 72 from both sides

72 - 72 + 3y = 60 - 72 72 cancels on the left

3y = - 12 Divie both sides by 3

3y/3 = -12/3 3 cancels on the left because 3/3 = 1

y = -4

Now plug your answer for y back into the first equation to get x.

x + y = -8

x + (-4) = - 8 Add 4 to each side

x - 4 + 4 = - 8 + 4 4 cancels on the left

x = -4

x = - 4 and y = - 4

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2 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

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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:

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:

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

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

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