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marin [14]
3 years ago
15

Which method would i use to solve ax^2+c=d? justify your answer.

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0

9514 1404 393

Answer:

  x = ±√((d -c)/a)

Step-by-step explanation:

The usual "solve for ..." approach works fine in this case.

  ax^2 +c = d

  ax^2 = d - c . . . . . subtract c

  x^2 = (d -c)/a . . . . divide by a

  x = ±√((d -c)/a) . . . take the square root

_____

When there are numbers, a graphical solution could work. It depends on whether an exact or approximate solution is required.

_____

<em>Comment on "solve for ..."</em>

When solving an equation for a specific variable, it is worthwhile to consider the Order of Operations. Look at what has been done to the variable, then undo those operations in reverse order. Here, the variable has been squared, the product multiplied by 'a', that product added to c. So, the "undo" consists of subtracting 'c', dividing by 'a', and taking the square root.

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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
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