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Marizza181 [45]
3 years ago
12

Which of the following characteristics does not describe a stock?

Mathematics
2 answers:
kipiarov [429]3 years ago
8 0

hi

it'b :  If you earn money from a coupon, it's mean that you  lentmoney to somenone.in  In stocks, there is no  coupon.  The annual meeting of stock holders will decide how much is  given for each share by considering annual profit made by the compagny.

WITCHER [35]3 years ago
6 0

Answer:

B. Money earned from coupon rate

Step-by-step explanation:

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16% of $36.48 is $5.84

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Cross multiply 16 and 36.48 to 583.68, and then I divided by 100, to get the percent of 5.8368, which I rounded up to 5.84%.
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Two lines are parallel to each other if they have the same slope. True or false?
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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 <em>E</em>_{i}, of the sample space <em>S,</em> given that another event <em>A</em> 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 <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

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

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = 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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Step-by-step explanation:

.............................

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