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jeka94
3 years ago
8

Which two of the following are applications of the Logarithmic model? *

Mathematics
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

Richter Scale and Acidity/Alkalinity

Step-by-step explanation:

College GPA is modelled by weighted means, Compound Interest uses a potential model, amortization schedule is based on succesions, whereas Richter Scale and Acidity/Alkalinity use logartihmic model.

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fiasKO [112]
5x15 is 75 and 7x6 is 42
8 0
2 years ago
Read 2 more answers
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
Need help on this . Explain how you got it ?
kvasek [131]
The goal is to find a number between 0.125 and 0.18
1/5 is 0.2
0.2, 1.6, 0.09 are not in the range. Therefore, the square root of 0.02 (0.1414) is the answer.
8 0
3 years ago
A population of 200 animals is decreasing by a rate of .96 per year . At this rate, in how many years will the population be les
mojhsa [17]
At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ

Now, we need to know when the population will be less than 170.

So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n

n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176
7 0
3 years ago
2) translate into an equation
Butoxors [25]
-4 - 7 divide 8n thats the answer
6 0
3 years ago
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