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kipiarov [429]
3 years ago
6

NEED ANSWER ASAP PLEASE WILL GIVE BRAINLIEST An integer n is divided by 2 and the quotient is an even integer. What does this te

ll you about n? Give an example. ​ ​
Mathematics
1 answer:
Nesterboy [21]3 years ago
4 0

Answer:

n is a multiple of 4

Step-by-step explanation:

Let's think about this.

Even numbers are basically numbers that, when divided by 2, get us an integer.

This means that n must be an even integer.

<em>However,</em>

Every other even number, when divided by 2, gets you an odd integer.

This means that every second even number, n \div 2 gets us an even number.

This also means that every time n \div 2 is an even number, n will be a multiple of 4 (as 2\cdot 2 = 4).

Test this out with any even number that's a multiple of 4 (all multiples of 4 are even numbers)

Hope this helped!

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In your calculator, input arcsin(7/12). Make sure that your calculator is in degree mode. The answer is 35.69 degrees.
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4 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

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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