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3 years ago

Plz help I'm and dead confused

Mathematics

1 answer:

boyakko [2]3 years ago

6 0

<h3>Answer: Choice D</h3>

Domain: {-3, -1, 2, 4, 6}

Range: {-2, 1, 2, 4, 6}

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Explanation:

The domain is the set of possible x values of a relation or function. To form the domain, we just list out the x coordinates of each point

Point A is located at (-3, 4) so the x coordinate x = -3 is part of the domain

Point B is at (-1,1) meaning x = -1 is also part of the domain

And so on. Doing this for all five points leads to the list of x coordinates {-3,-1,2,4,6} which is the domain.

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The range is a similar story, but we look at the y coordinates now. Point A has y coordinate 4, so y = 4 is part of the range.

Point B has y coordinate y = 1, so that is also part of the range.

The range is {-2, 1, 2, 4, 6} representing all of the possible y values of this relation.

With any set, the order does not matter. Though convention is to list the items from smallest to largest.

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What is 1 and 1/4 minus 3/8?

matrenka [14]

Answer: 7/8

Step-by-step explanation:

You can change all the fractions to have equivalent denominators.

The Least Common Denominator of 1 1/4 and 3/8 is 8. Now we can change the fractions.

1 1/4 = 1 2/8 = 10/8.

3/8 = 3/8.

Now we can subtract. Subtract the numerators of the first fraction from the numerators of the second fraction, and the denominators oft he first fraction to the denominators of the second fraction.

10/8 - 3/8 = 7/8.

Therefore, 1 1/4 - 3/8 = 7/8; the fourth answer.

6 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

ANSWER FOR BRAINLIEST :)

Tju [1.3M]

Answer:

simplified : k/2 - 3/5

Step-by-step explanation:

hope this helps <3

8 0

3 years ago

Read 2 more answers