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LuckyWell [14K]
2 years ago
10

There are 700 people at a funfair.

Mathematics
2 answers:
qwelly [4]2 years ago
5 0

Answer:

Answer is in Picture. Explanation as well

Eddi Din [679]2 years ago
5 0
700/7 = 100
100 * 4 = 400
There are 400 people are boys.

700/10 = 70

400 + 70 = 470
700 - 470 = 230
There are 230 adults at the funfair.

I hope I helped. I appreciate Brainliest
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A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade
Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

P(x\geq1)=1-P(x

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly  n= 5

Number of blades in an assembly dull is M  = 10.

The probability mass function is,

P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

P(x\geq 1)=1- P(x

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly  N = 48

Number of blades selected at random from the assembly  N = 5

Number of blades in an assembly dull is  M = 10

From the information,

The probability that assembly is replaced (P)  is 0.7069.

The probability that assembly is not replaced is (Q)  is,

q=1-p\\= 1-0.7069= 0.2931

The geometric probability mass function is,

P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)

The probability that assembly is replaced no replaced until the third day of evaluation is,

P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly   N = 48

Number of blades selected at random from the assembly  n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 2.

P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 6.

P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

P(x\geq 1)=1- P(x

 

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0
3 years ago
Question 3. Solve each equation given<br>below.<br>5 x - 4=31​
harkovskaia [24]

Answer:

x = 7

Step-by-step explanation:

5x-4 = 31

5x = 31+4

5x = 35

x = 7

3 0
2 years ago
Can some one answer quickly please
denis-greek [22]

Answer:

the answer is A -1/3 x

Step-by-step explanation:

hope this helps :)

3 0
3 years ago
An employee has been stealing funds from the company for years. Which of the following control methods would have most likely un
nasty-shy [4]

Answer:

Complete a bank reconciliation

Step-by-step explanation:

A bank reconciliation is the process where the financial records of the bank is compared with the firm's financial record. If an employee has been stealing funds from the company, there would be a discrepancy between the figures provided by the record.

In a bid to know the reason for the discrepancy, the theft would be detected

8 0
2 years ago
A. Steven wrote an expression with 2 terms. The second term, which is the whole number 7, is subtracted from the first term. The
bulgar [2K]

Answer:

Part a)

Steve's expression is 2(2x+5)^{2}-7

For x=3, the expression is equal to 235

Part b)

Jasmine's expression is

(3x^{2})+(25x)

For x=2, the expression is equal to 62

Step-by-step explanation:

Part a)

Let

n-----> the first term

The expression is

(n-7)

n=2(2x+5)^{2}

substitute

Steve's expression is

2(2x+5)^{2}-7

Evaluate for x=3

2(2(3)+5)^{2}-7=242-7=235

Part b)

Let

n-----> the first term

Jasmine's expression is

(3x^{2})+(25x)

Evaluate for x=2

(3(2)^{2})+(25(2))=12+50=62

4 0
3 years ago
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