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Vikentia [17]
3 years ago
13

A restaurant has 1,996 forks 1,745 knives and 2,116 spoons. The owner wants to have 2,000 of each utensil. How many additional f

orks. And knives are needed ? How many extra spoons are there?
Mathematics
2 answers:
Assoli18 [71]3 years ago
6 0

The restaurant needs 4 more forks, 255 more knives and there are 116 extra spoons

a_sh-v [17]3 years ago
4 0

2000-1996=4 forks

2000-1745=255 knives

since there are extra spoons it is reversed-->2116-2000=116 spoons

the owner needs 4 forks,255 knives and there are 116 extra spoons


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If x2+y2=40 and xy=5 then find(x+y)2
VMariaS [17]

Answer:

(x + y)^2 = 45

Step-by-step explanation:

(x + y)^{2} = x^{2}  +xy + y^{2}

             = 40 + 5 = 45

8 0
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Solve: 2x ≤ -2/3 (4x + 4)
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Answer: x ≤ 1

reason:

first, multiply 4x and 4 by -2/3, which leaves you with the equation 2x ≤ -8/3x + 8/3
then, you add 8/3x to both sides, making it 4 2/3x ≤ 8/3
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Make cos C the subject of the formula<br> C?= a? + b? - 2ab cos C
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2 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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