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laiz [17]
3 years ago
8

Marques earned a grade of 54% on his multiple choice math final that had a total of 200 problems. How many problems on the final

exam did Marques answer correctly?
Mathematics
1 answer:
seraphim [82]3 years ago
8 0

Answer:

108

Step-by-step explanation:

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Because 40% of 45 is 18

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2 years ago
Triangle N Q L has centroid S. Lines are drawn from each point through the centroid to the midpoint of the opposite side to form
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36

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The smaller triangle was dilated to form the larger triangle what is the value of x
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i think c which is 14

3 0
3 years ago
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
The transitive property of angle congruence states that if ________ ≅ ∠QRS and ∠QRS ≅ ∠PQR, then ∠XYZ ≅ ∠PQR . A) ∠XQR B) ∠QRS C
djverab [1.8K]

Answer:

The answer is ∠XQR

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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