Hit me up. Bet i snap back. Ianh

What are the key features of a graph of a quadratic equation with only complex solutions

dybincka [34]

Answer:

If D< 0, the roots of the quadratic equation will b complex.

Step-by-step explanation:

Let $ax^{2} + bx + c =0$ is the given quadratic equation.

So, if the Discriminant (D) = $b^{2} - 4ac < 0$ , the the equation will have only complex solutions.

Complex roots are represented in the form of $(x \pm iy)$

1. x represents the X axis (real) co ordinate,

y represents the Y axis (imaginary) coordinate

2. Also, the graph of the with equation D < 0 NEVER CROSSES the X - axis, as there are no real roots of the equation, only complex roots.

3. Here. roots are always identical but are opposite in signs. Such pair or roots with opposite sign are called CONJUGATE PAIRS.

4 0

3 years ago

A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic

goldfiish [28.3K]

Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event E $_{i}$ , of the sample space S, given that another event A has already occurred is:

$P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}$

The law of total probability states that, if events E₁, E₂, E₃... are parts of a sample space then for any event A,

$P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}$

Denote the events as follows:

X = fetus have a chromosome abnormality.

Y = the test is positive

The information provided is:

$P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85$

Using the above the probabilities compute the remaining values as follows:

$P(X^{c})=1-P(X)=1-0.04=0.96$

$P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10$

$P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15$

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

$P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})$

$=(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82$

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}$

$=\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}$

$=0.20$

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

6 0

3 years ago

g Bonus: Assume that among the general pediatric population, 7 children out of every 1000 have DIPG (Diffuse Intrinsic Pontine G

patriot [66]

Answer:

0.9586

Step-by-step explanation:

From the information given:

7 children out of every 1000 children suffer from DIPG

A screening test designed contains 98% sensitivity & 84% specificity.

Now, from above:

The probability that the children have DIPG is:

$\mathbf{P(positive) = P(positive \ | \ DIPG) \times P(DIPG) + P(positive \ | \ not \DIPG)\times P(not \ DIPG)}$ $= 0.98\imes( \dfrac{7}{1000}) + (1-0.84) \times (1 - \dfrac{7}{1000})$