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3 years ago

8

How much will Bob need to save each month if he wants to buy a $30,000 car with cash in 5 years? He can earn a nominal interest

rate of 10% compounded monthly.
a) $2.50
b) $250.00
c) $25.00
d) $1,862.76

1 answer:

NeX [460]3 years ago

7 0

Answer:

B

Step-by-step explanation:

C is too little and D is too much

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Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

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

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

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3 years ago

How is (y=15x) a proportional relationship?

Elina [12.6K]

Answer:

because y and x are equivalent and equal 15

Step-by-step explanation:

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3 years ago

Set up the trig equation with the correct values plugged in that you would use to solve for h. (You do not need to solve)

Usimov [2.4K]

From the reference of the 18 degree angle, 'h' is the opposite side and 100 is the adjacent side.

The trig ratio which uses both the opposite and adjacent sides is the tangent.

tan(18) = opp/adj = h/100

Trig equation:
tan(18) = h/100

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3 years ago

Choose the congruence theorem that you would use to prove the triangles congruent.

LekaFEV [45]

The correct answer is SSS

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2 years ago

What is the value of x^3*y^1 when x=5 and y=10

Varvara68 [4.7K]

Answer:

1250

Step-by-step explanation:

x^3*y^1

Let x = 5 and y = 10

5^3* 10^1

PEMDAS says powers first

125 * 10

Then multiply

1250

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3 years ago