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mote1985 [20]
3 years ago
8

Finish the chart for y= 1/2x

Mathematics
2 answers:
bazaltina [42]3 years ago
7 0
It would be -2=-1, 0=0 , 2=1

Hope this helps

Have a great day/night
11Alexandr11 [23.1K]3 years ago
4 0

Answer:

So we know that y=1/2x. So we apply this equation on each x.

x=-2, so y=1/2*-2 which is -1. so y = -1

x=0, so y = 1/2 * 0 which is 0. so y=0

x=2, so y = 1/2*2 which is 1. so y=1

Hope this helps :))

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Jemila has $10, $5, and $1 bills in her piggy bank. Their total value is $63. If she has two more $5 bills than $10 bills, and t
earnstyle [38]

Answer:

She has 3 $10 bills, 5 $5 bills and 8 $1

3 times 10= $30

5 times 5 = $25

8 times 1 = $8

$30 + $25 + $8 = $63

4 0
3 years ago
Read 2 more answers
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

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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