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3 years ago

If x is a positive integer, for how many different values of x is sqrt48/x a whole number?

Mathematics

1 answer:

slava [35]3 years ago

6 0

Answer:

The only possible values of x that fulfill with both conditions are 3, 12 and 48.

Step-by-step explanation:

1. What are the positive integers? Also know as the whole numbers or the counting numbers. They start from 1, and continue to 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, etc.

2. What are the whole numbers? The definition of whole numbers is that they are the number with no fractions or decimals.

3. According to those definitions, x must be a positive integer and the result of √48/x, must be a whole number.

4. For fulfilling the first condition, x must be only a value from 1 to 48.

5. For fulfilling the second condition, the result of √48/x must be a whole number. It means it must be 1, 2, 3, 4, 5 or 6. From 7 is not possible to continue because those numbers are bigger than 48, for example √49, √64, √81, √100 and so on, and it would be impossible to fulfill with the first condition.

6. So, we have just 6 possible options (√48/x = 1, √48/x = 2, √48/x = 3, √48/x = 4, √48/x = 5 and √48/x =6). Let's review in detail one by one.

√48/x = 1, Is there any value of x for fulfilling both conditions? Yes, it's. It's 48. 

√48/x = 1 and let's replace x by 48.

√48/48 = √1 = 1

√48/x = 2, Is there any value of x for fulfilling both conditions? Yes, it's. It's 12. 

√48/x = 2 and let's replace x by 12.

√48/12 = √4 = 2

√48/x = 3, Is there any value of x for fulfilling both conditions? No, it isn't.

There isn't any whole number that allow us to fulfill with this: √48/x = √9 = 3

√48/x = 4, Is there any value of x for fulfilling both conditions? Yes, it's. It's 3.

√48/x = 4 and let's replace x by 3.

√48/3 = √16 = 4

√48/x = 5 Is there any value of x for fulfilling both conditions? No, it isn't.

There isn't any whole number that allow us to fulfill with this: √48/x = √25 = 5

√48/x =6 Is there any value of x for fulfilling both conditions? No, it isn't.

There isn't any whole number that allow us to fulfill with this: √48/x = √36 = 6.

The only possible values of x that fulfill with both conditions are 3, 12 and 48.

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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

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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

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