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Tju [1.3M]
4 years ago
7

There are 2 leaves along 3 in. of an ivy vine. There are 14 leaves along 15 in. of the same vine. Which equation models the numb

er of leaves y along x in. of vine?
Mathematics
2 answers:
Hunter-Best [27]4 years ago
8 0
Doing a bit of guess and check, we can notice that the amount of leaves is one less than the number of inches of vine. Writing it out, we get x-1=y (the amount of inches-1=the amount of leaves)
Ugo [173]4 years ago
8 0

The answer on Edgenuity2020 is b. 5 leaves

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4 times 80 hope I helped
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3 years ago
Suppose the first term of a geometric sequence is 3 and the fourth term is 192. What is the common ratio of the sequence? Write
gayaneshka [121]
An=A1 r ^(n-1)
192= 3 r ^(4-1)
192 = 3 r^3
64 = r^3
r = 4

equation:
An=3 (4)^(n-1)

6th term:
An= 3 (4)^(6-1)
An= 3072

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3 0
3 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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3 years ago
ILL GIVE BRAINLY HELP 20 POINTS!!!
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Answer:

13.6

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Daniel makes 16 more muffins than kris.
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