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Kruka [31]
3 years ago
5

A person has two times as many nickels as quarters. if the total face value of these coins is $4.55, how many of each type of co

in does this person have?
Mathematics
1 answer:
Sloan [31]3 years ago
8 0
<span>Let
He has = x quarters
He has = 2x nickels 

Total Amount = $3.50 or 3.50*100 = 350 cents 

x*25+2x*5=350
25x+10x=350
35x=350

35x/35=350/35
x=10 

He has = x = 10 quarters 
He has = 2x = 2*10 = 20 nickels 

Check
====== 
10*25 + 5*20 = 350
250+100=350
350=350</span>
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Answer:

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Step-by-step explanation:

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3 years ago
An electrical rm manufactures light bulbs that have a life span that is approximately normally distributed. The population stand
stira [4]

Answer:

The 't' test statistic = 1.46 < 1.69

The test of hypothesis is H 0 :μ = 800 is accepted

A sample of 30 bulbs are found came from average µ= 800

Step-by-step explanation:

Step 1:-

Given population of mean μ = 800

given size of small sample n =30

sample standard deviation 'S' = 45

Mean value of the sample χ = 788

Null hypothesis H_{0} =µ  =800

alternative hypothesis  H_{1} = µ ≠ 800

<u>Step 2</u>:-

The 't' test statistic t =  \frac{x-μ}{\frac{sig}{\sqrt{n} } }

                             t = \frac{788-800}{\frac{45}{\sqrt{30} } }

                       t = \frac{12}{8.215} = 1.4607

<u>Step 3</u>:-

The degrees of freedom γ = n-1 = 30-1 =29

From "t" value from table at 0.05 level of significance ( t = 1.69)

The calculated value t = 1.4607 < 1.69

Therefore The null hypothesis H_{0} ' is accepted.

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A sample of 30 bulbs are came found from average µ= 800

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4 years ago
3. Assume that the likelihood of a child under the age of ten watching PBS is 0.76. Three children are
natta225 [31]

Using the binomial distribution, the probabilities are given as follows:

a) 0.4159 = 41.59%.

b) 0.5610 = 56.10%.

c) 0.8549 = 85.49%.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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.

For this problem, the values of the parameters are:

n = 3, p = 0.76.

Item a:

The probability is P(X = 2), hence:

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

P(X = 2) = C_{3,2}.(0.76)^{2}.(0.24)^{1} = 0.4159

Item b:

The probability is P(X < 3), hence:

P(X < 3) = 1 - P(X = 3)

In which:

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

P(X = 3) = C_{3,3}.(0.76)^{3}.(0.24)^{0} = 0.4390

Then:

P(X < 3) = 1 - P(X = 3) = 1 - 0.4390 = 0.5610 = 56.10%.

Item c:

The probability is:

P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4159 + 0.4390 = 0.8549

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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Answer:

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Step-by-step explanation:

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Answer:

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