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Hoochie [10]
3 years ago
10

Calculate the area of the circle with radius of 4.5 leave Answer in terms of pi

Mathematics
1 answer:
aleksklad [387]3 years ago
8 0
A=pi r^2
= 20.25 pi is the answer
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The measure of central angle YCZ is 80 degrees
Vera_Pavlovna [14]

Answer:

The answer is 36 units2

Step-by-step explanation:

I took the test so I’m right

6 0
3 years ago
Read 2 more answers
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 help expand the expression using pascal's triangle<br><br> (1-2x)^4
aliya0001 [1]
(1 - 2x)⁴
(1 - 2x)(1 - 2x)(1 - 2x)(1 - 2x)
[1(1 - 2x) - 2x(1 - 2x)][1(1 - 2x) - 2x(1 - 2x)]
[1(1) - 1(2x) - 2x(1) - 2x(-2x)][1(1) - 1(2x) - 2x(1) - 2x(-2x)]
(1 - 2x - 2x + 4x²)(1 - 2x - 2x + 4x²)
(1 - 4x + 4x²)(1 - 4x + 4x²)
1(1 - 4x + 4x²) - 4x(1 - 4x + 4x²) + 4x²(1 - 4x + 4x²)
1(1) - 1(4x) + 1(4x²) - 4x(1) - 4x(-4x) - 4x(4x²) + 4x²(1) - 4x²(4x) + 4x²(4x²)
1 - 4x + 4x² - 4x + 16x² - 16x³ + 4x² - 16x³ + 16x⁴
1 - 4x - 4x + 4x² + 16x² + 4x² - 16x³ - 16x³ + 16x⁴
1 - 8x + 24x² - 32x³ + 16x⁴
5 0
3 years ago
Read 2 more answers
Evaluate q^2-5q+4, when q equals -2
Anastasy [175]

Answer:

18

Step-by-step explanation:

q^2 - 5q +4

Let q = -2

(-2)^2 - 5(-2) +4

4  + 10 +4

18

8 0
3 years ago
What is the 12th term of the sequence 3,6,12,24....
Sati [7]
a_{0}=3

r=common ratio: 12/6=2; 6/3=2

r=2, a_{n}=2^{n-1}*3

plug in 12 for n:

a_{12}=2^{12-1}*3

a_{12}=2^{11}*3

a_{12}=2048*3

a_{12}=6144

The twelfth term in the sequence is 6144


 




7 0
3 years ago
Read 2 more answers
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