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3 years ago

HELP ME PLSSSSSSSSSSSSSSSSSSSSSSS I GIVE BRAINLIEST

Mathematics

2 answers:

AURORKA [14]3 years ago

6 0

Answer:

the answer is c

Step-by-step explanation:

CAN I PLEASE GET BRAINLESS

postnew [5]3 years ago

4 0

Answer:

Answer is 5. Hope you like answer

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2x³+16y³= 2(x³+8y³)

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What is the simplified form of the quantity of x plus 7, all over the quantity of x plus 3 + the quantity of x minus 4, all over

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$\frac{3x + 21}{4x + 5}$

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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.

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4 years ago

A box is filled with 3 yellow cards , 4 green cards. A card is chosen at random from the box. What is the probability that the c

nasty-shy [4]

Answer:

3/7

Step-by-step explanation:

total cards 3+4 = 7

Not green = 7-4 = 3 cards

P ( not green )= not green cards / total = 3/7

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Artemon [7]

Answer:

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