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3 years ago

Question:

Mathematics

1 answer:

faltersainse [42]3 years ago

3 0

Answer:

The probability that a ship that is declared defecive is sound is 0.375

Step-by-step explanation:

Let P(A|B) denote the conditional probability of A given B. We will make use of the equation

P(A|B) = P(A) × P(B|A) / P(B)

We have the probabilities:

P(Declared Defective (detected) | Defective) = 0.95

P(not Detected | Defective) = 1-0.95=0.05

P(Declared Sound | Sound) = 0.97

P(Declared Defective |Sound) = 1-0.97=0.03

P(Defective)=0.05

P(Sound)= 1- 0.05 = 0.95

We can calculate:

P(Declared Defective)= P(Detected | Defective)×P(Defective) + P(Declared Defective |Sound) ×P(Sound) = 0.95×0.05 + 0.03×0.95=0.076

P(S | Declared Defective) =

(P(Sound) × P(Declared Defective | Sound)) / P(Declared Defective)

=0.95×0.03 /0.076 =0.375

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An auto insurance company has 10,000 policyholders. Each policy is defined as: young or old male or female; and married or singl

Blababa [14]

Answer:

880

Step-by-step explanation:

This problem a bit annoying, but let's classify the numbers and see where we can get from there:

Total = 10,000

Young = 3,000

Old = 7,000 (not young defined as 10,000 - 3,000)

Male = 4,600

Female = 5,400

Married = 7,000

Single = 3,000

Young Males = 1,320

Young Females = 3,000 (young) - 1,320 = 1,680

Married Males = 3,010

Married Females = 7,000 - 3,010 = 2,990

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Young Married Males = 600

Young Married Females = 1,400 - 600 = 800

Young Female Single = ?

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6 0

3 years ago

Suppose yoko borrows $3500 at an interest rate of 4% compounded each year Find the amount owed at the end of 1 year

olchik [2.2K]

Amount owed at the end of 1 year is 3640

<h3>Solution:</h3>

Given that yoko borrows $3500.

Rate of interest charged is 4% compounded each year

Need to determine amount owed at the end of 1 year.

In our case :

Borrowed Amount that is principal P = $3500

Rate of interest r = 4%

Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1

Formula for Amount of compounded yearly is as follows:

$A=p\left(1+\frac{r}{100}\right)^{n}$

Where "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula we get

$\mathrm{A}=3500\left(1+\frac{4}{100}\right)^{1}$

$\begin{array}{l}{A=\frac{3500 \times 104}{100}} \\\\ {A=35 \times 104} \\\\ {A=3640}\end{array}$

Hence amount owed at the end of 1 year is 3640

5 0

3 years ago

Solve using the substitution method. Show your work. If the system has no solution or an infinite number of solutions, state thi

vladimir2022 [97]

The answer is x = 10, y = 10.

Step 1: rearrange the second equation for y.
Step 2: substitute y from the second equation into the first equation.
Step 3. Calculate y.

Step 1.
The second equation is: 6x + 3y = 90
Divide both sides of the equation by 3:
(6x + 3y)/3 = 90/3
6x/3 + 3y/3 = 30
2x + y = 30
Rearrange the equation:
y = 30 - 2x

Step 2.
Substitute y from the second equation (y = 30 - 2x) into the first equation:
15x + 9y = 240
15x + 9(30 - 2x) = 240
15x + 270 - 18x = 240
15x - 18x = 240 - 270
-3x = -30
x = -30/-3
x = 10

Step 3.
Since y = 30 - 2x and x = 10, then:
y = 30 - 2 * 10
y = 30 - 20
y = 10

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3 years ago