120,395 as a prrcent

Among 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in three different pacemakers randomly selected from this batch of 8834 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

Answer:

P(All three are not caused by firmware) = 83.84%

Probability that the entire batch will be accepted = 0.8384

Step-by-step explanation:

We are told that out of the 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware.

Thus,

Cases not caused by firmware = 8834 - 504 = 8330

So, probability of the first case not being affected by firmware is;

P(first case not caused by firmware) = 8330/8834

Also,

Probability of second case not being affected by firmware is given as;

P(second case not caused by firmware|first case not affected by firmware) = 8329/8833

Similarly,

Probability of third case not being affected by firmware is given as;

P(third case not caused by firmware|first and second not caused by firmware) = 8328/8832

Now, looking at the 3 Probabilities gotten, it is obvious that the events are not independent because the probability of occurence of one event depends on the probability of occurence of the other event.

Thus, we will make use of the general multiplication rule which is;

P(A & B) = P(B) × P(A|B)

Thus;

P(All three not caused by firmware) = P(first case not caused by firmware) × P(second case not caused by firmware|first case not affected by firmware) × P(third case not caused by firmware|first and second not caused by firmware)

Plugging in the relevant values, we have;

P(All three not caused by firmware) = (8330/8834) × (8329/8833) × (8328/8832)

P(All three are not caused by firmware) = 0.83840506679 ≈ 83.84%

6 0

3 years ago

Carlos deposited $7,924 into a savings account 30 years ago. The account has an interest rate of 4.6% and the balance is current

exis [7]

Answer:

n = 1, this means the interest compounds ANNUALLY.

Step-by-step explanation:

Carlos deposited $7,924 into a savings account 30 years ago. The account has an interest rate of 4.6% and the balance is currently $30,541.83. How often does the interest compound?

Compound Interest Formula

: A = P(1 + r/n)^nt

A = Amount after time t

P = Principal (Initial Amount Invested)

r = Interest rate

n = Number of times the interest is compounded

t = time in years

A = $30,541.83

r = 4.6% = 0.046

t = 30

P = $7,924

Hence,

$30,541.83 = $7924(1 + 0.046/n)^30n

Divide both sides by 7924

$30,541.93/$7924 = (1 + 0.046/n)^30n

$30,541.93/$7924 = (n + 0.046/n)^30n

3.8543576477 = (n + 0.046/n)^30n

We take the logarithm of both sides

log 3.8543576477 = log (n + 0.046/n)^30n

Solving for n,

n = 1

Therefore, from the calculation above, since n = 1, this means the interest compounds ANNUALLY.

6 0

2 years ago