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Olenka [21]
3 years ago
14

You are creating an emergency fund and decide to place $415.00/month in an account that earns a 3.75 APR. How much interest accr

ues by the end of the first full month?
Mathematics
1 answer:
hoa [83]3 years ago
4 0

At the end of the first full month the interest acquires is $ 1.30.

<u>Step-by-step explanation:</u>

Since we are looking for the interest for full month, we have to divide the APR by 12.

Because APR is the interest amount for the full year.

So 3.75 % = 3.75/100 = 0.0375

Now multiplying, we will get

415 × 0.0375/12 = $ 1.30

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Whitepunk [10]

Answer:

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Step-by-step explanation:

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8 0
2 years ago
The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
3 years ago
Solve this inequality for x. 81-1 1/5 x &lt;55
katen-ka-za [31]

Answer:

A. x > 21 2/3

Step-by-step explanation:

x is greater than 21 and two thirds

i did the math :]

4 0
3 years ago
I need help with Geometry
Licemer1 [7]

ednocrkdlwmqcw e9rbeuopgjkmr qoejffwkf ,rwfrwfjr nmwrfHELP ME iuehf[eoiffkjkmfeiufkwjmfwufuStep-by-step explanation:

5 0
3 years ago
HELP PLEASE DUE IN 3 MINUTES
Arisa [49]
Answer: 6

Steps:
7 + 11 + 0 = 18
18/3 = 6
3 0
3 years ago
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