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3 years ago

Help qwq im sorta confused-

Mathematics

1 answer:

zloy xaker [14]3 years ago

7 0

Answer:

it should be 1,632.....

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(y-1)(y^2+2y+1) Simplify please

sashaice [31]

Do like this

(y - 1)(y² + 2y + 1)

y³ + 2y² + y - y² - 2y² - 1

y³ - 1

This is end result

3 0

3 years ago

A public health researcher examines the medical records of a group of 937 men who died in 1999 and discovers that 210 of the men

Julli [10]

Answer:

The correct option is (B) 0.173.

Step-by-step explanation:

The law of total probability states that:

$P(A)=P(A\cap B)+P(A\cap B^{c})$

The conditional probability of an event A given that another event B has already occurred is:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

Then the probability of intersection of A and B is:

$P(A\cap B)=P(A|B)\cdot P(B)$

Denote the events as follows:

H = a man died from causes related to heart disease.

X = a man had at least one parent who suffered from heart disease

The information provided is:

$P(H)=\frac{210}{937}\\\\P(X)=\frac{312}{937}\\\\P(H|X)=\frac{102}{312}$

The probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered from heart disease is, $P(H|X^{c})$ .

Compute the value of $P(H|X^{c})$ as follows:

$P(H)=P(H|X)\cdot P(X)+P(H|X^{c})\cdot P(X^{c})$

$\frac{210}{937}=[\frac{102}{312}\cdot \frac{312}{937}]+[P(H|X^{c})\cdot (1-\frac{312}{937})]\\\\\frac{210}{937}-\frac{102}{937}=[P(H|X^{c})\cdot \frac{625}{937}]\\\\\frac{108}{937}=P(H|X^{c})\cdot \frac{625}{937}\\\\P(H|X^{c})=\frac{108}{937}\times \frac{937}{625}\\\\P(H|X^{c})=0.1728\\\\P(H|X^{c})\approx 0.173$

Thus, the probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered from heart disease is 0.173.

The correct option is (B).

4 0

3 years ago

What is the volume of the triangular prism below 7.8 in 6 in 5 in 11.5 in

Zarrin [17]

Answer:

D. 172.5 inches cubed

Step-by-step explanation:

Have a good day :)

8 0

3 years ago

The heights of a random sample of students in stat 1350 were recorded in inches. the heights were then converted to the metric s

Kisachek [45]

There is no difrence between stat and cenrsimete

7 0

3 years ago

What is the area of the rhombus 10in and 5in

PolarNik [594]

Answer:25 in^2

Step-by-step explanation:

d1=10 d2=5

area of rhombus=( d1 x d2)/2

Area of rhombus=(10x5)/2

area of rhombus=50/2

Area of rhombus=25 in^2

5 0

3 years ago