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SVETLANKA909090 [29]
2 years ago
9

What is the greatest common factor of 120 and 168

Mathematics
2 answers:
GaryK [48]2 years ago
5 0

Answer:

24

Step-by-step explanation:

gcf for that is 24 I think

Evgen [1.6K]2 years ago
5 0

Answer:

24

Step-by-step explanation:

120/24 = 5

168/24 = 7

5 and 7 do not have any common factors, hence they cannot be reducible.

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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
Please Answer!!!!
ExtremeBDS [4]

First mug holds the most

<em><u>Solution:</u></em>

Given that,

You are choosing between two mugs

<em><u>The volume of cylinder is given as:</u></em>

V = \pi r^2h

Where,

r is the radius and h is the height

<em><u>One has a base that is 5.5 inches in diameter and a height of 3 inches</u></em>

radius = \frac{diameter}{2}

Therefore,

r = \frac{5.5}{2}\\\\r = 2.75

Also, h = 3 inches

<em><u>Thus volume of cylinder is given as:</u></em>

V = \pi \times 2.75^2 \times 3\\\\V = 3.14 \times 22.6875\\\\V = 71.23875 \approx 71.24

Thus first mug holds 71.24 cubic inches

<em><u>The other has a base of 4.5 inches in diameter and a height of 4 inches</u></em>

Radius = \frac{4.5}{2}\\\\r = 2.25

h = 4 inches

Therefore,

V = 3.14 \times 2.25^2 \times 4\\\\V = 3.14 \times 20.25\\\\V = 63.585

Thus the second mug holds 63.585 cubic inches

On comparing, volume of both mugs,

Volume of first mug > volume of second mug

First mug holds the most

8 0
3 years ago
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