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BabaBlast [244]
3 years ago
12

Alicia buys a 5 pound bag of rocks for a fish tank she uses 1 1/8 pounds for a small fish bowl how much is left

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Alicia would have 3 7/8 pound of rocks leftover

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It’d be 0.57, hope this helps.
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How do you solve the equation 3w-(-4)=8-w
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3w-(-4)=8-w
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Reduce 27/42 to the lowest term
wel
So you need to find a number that goes into both of these numbers:
3 goes into both, so divide both by 3
27/3 = 9
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A triangle has vertices (2, 1), (5, 1), and (5,5). What is the length of the longest side of the triangle?
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8 0
3 years ago
The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.
mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0
3 years ago
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