What is the final amount of a $300 investment earning 4.5% interest compounded quarterly for 3 years?

1 answer:

wel2 years ago

7 0

The formula A=P(1+(r/n))^nt should be followed when answering
300=P
R=0.045
N=4
T=3
300(1+(0.045/4))^4x3
Answer is=343.102

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the cost of a phone call lasting 3 minutes 30 seconds was 52.5 cents . at this rate what was the cost of a call lasting 5 minute

Effectus [21]

Answer:

80 cents

Step-by-step explanation:

The easiest place to start for this is to calculate how much it costs per minute of call time. To do this, if we know that it costs 52.5 cents to call for 3.5 minutes, we can divide those two numbers to get how much it costs per minute.

52.5/3.5 = 15

If it costs 15 cents per minute, and we want to know how much it would cost to call for 5.33 (5 and 1/3 of a minute), then we multiply our 15 cents a minute by the number of minutes to get the final cost.

15 x 5.33 = 79.99

Because we can't have 99/100 cents, rounding up to 80 is important to get a proper answer.

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3 years ago

It has a time to failure distribution which is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehi

Triss [41]

Answer:

The designed life should be of 21,840 vehicle miles.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.

This means that $\mu = 35000, \sigma = 7000$

Find its designed life if a .97 reliability is desired.

The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.

$Z = \frac{X - \mu}{\sigma}$

$-1.88 = \frac{X - 35000}{7000}$