Thank you, have a great day.

I just need an explanation on why I was wrong and why the correct answer is correct

Alja [10]

Answer:

B 1

Step-by-step explanation:

Since the divisor is in the form of x - c, use what is called Synthetic Division. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

2| -2 1 5 0 4 1

↓ -4 -6 -2 -4 0

_________________

-2 -3 -1 -2 0 1→ -2x⁴ - 3x³ - x² -2x + [x - 2]⁻¹

You start by placing the c in the top left corner, then list all the coefficients of your dividend [-2x⁵ + x⁴ + 5x³ + 4x + 1]. You bring down the original term closest to c then begin your multiplication. Now depending on what symbol your result is, tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that -2 in your quotient can be a -2x⁴, and the -3 [x³] follows right behind it, then 1 [-x²], -2[x], and finally, [1\x - 2] (remainder is 1, so set it over your denominator, which is the divisor), giving you the other factor of -2x⁴ - 3x³ - x² -2x + [x - 2]⁻¹.

I am joyous to assist you anytime.

** $(x - 2)^{-1} = \frac{1}{x - 2}$

5 0

3 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:

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

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

Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05

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

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

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