Casey Bought 9 Tickets To A Concert The Total Charge Was $104 Including A $5 Service Charge

What is the a3+b3 equal to? verify by an activity

uranmaximum [27]

a³ + b³ = ( a + b ) ( a² - a b + b³ )
For example:
a = 1, b = 2
a³ + b³ = 1³ + 2³ = 1 + 8 = 9
( a + b ) ( a² - a b + b² ) = ( 1 + 2 ) ( 1² - 1 * 2 + 2² ) =
= 3 * 3 = 3
The formula is true.

3 0

2 years ago

Suppose you can afford only $200 a month in car payments and your best loan option is a 60-month loan at 3%. How much money coul

g100num [7]

Answer:

$11,130.47

Step-by-step explanation:

The amortization formula can be used. It tells you the monthly payment amount A for some principal P, interest rate r, and n payments.

A = P(r/12)/(1 -(1 +r/12)^(-n))

Filling in your values, we get ...

200 = P(.03/12)/(1 -(1 +.03/12)^-60) = P(.0025)/(1 -1.0025^-60)

P = 200(1 -1.0025^-60)/.0025 ≈ 200×55.6523577

P ≈ 11,130.47

The present value of the loan is $11,130.47.

8 0

2 years ago

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:

B = bugs are present in a computer program.

D = 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 E given that another event X 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.

4 0

2 years ago

there are 3 liters of apple juice at a school party. 10 students want to drink all the apple juice, Andi all want to get exactly

pochemuha

Answer:

0.3 L

Step-by-step explanation:

3 L / 10 students = 3/10 = 0.3 Liters per student

4 0

2 years ago

Jeff is baking a cake. The recipe says that he has to mix 32 grams of vanilla powder to the flour. Jeff knows that 1 cup of that

Sedaia [141]